The Quantum Mechanics Solver
The Ecole Polytechnique, one of France’s top academic institutions, has a longstanding tr...

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The Quantum Mechanics Solver

The Ecole Polytechnique, one of France’s top academic institutions, has a longstanding tradition of producing exceptional scientiﬁc textbooks for its students. The original lecture notes, the Cours de l’Ecole Polytechnique, which were written by Cauchy and Jordan in the nineteenth century, are considered to be landmarks in the development of mathematics. The present series of textbooks is remarkable in that the texts incorporate the most recent scientiﬁc advances in courses designed to provide undergraduate students with the foundations of a scientiﬁc discipline. An outstanding level of quality is achieved in each of the seven scientiﬁc ﬁelds taught at the Ecole: pure and applied mathematics, mechanics, physics, chemistry, biology, and economics. The uniform level of excellence is the result of the unique selection of academic staff there which includes, in addition to the best researchers in its own renowned laboratories, a large number of world-famous scientists, appointed as part-time professors or associate professors, who work in the most advanced research centers France has in each ﬁeld. Another distinctive characteristic of these courses is their overall consistency; each course makes appropriate use of relevant concepts introduced in the other textbooks. This is because each student at the Ecole Polytechnique has to acquire basic knowledge in the seven scientiﬁc ﬁelds taught there, so a substantial link between departments is necessary. The distribution of these courses used to be restricted to the 900 students at the Ecole. Some years ago we were very successful in making these courses available to a larger French-reading audience. We now build on this success by making these textbooks also available in English.

Jean-Louis Basdevant Jean Dalibard

The Quantum Mechanics Solver How to Apply Quantum Theory to Modern Physics

Second Edition

With 59 Figures, Numerous Problems and Solutions

ABC

Professor Jean-Louis Basdevant

Professor Jean Dalibard

Department of Physics Laboratoire Leprince-Ringuet Ecole Polytechnique 91128 Palaisseau Cedex France E-mail: [email protected] polytechnique.edu

Ecole Normale Superieure Laboratoire Kastler Brossel rue Lhomond 24, 75231 Paris, CX 05 France E-mail: [email protected]

Library of Congress Control Number: 2005930228 ISBN-10 ISBN-13 ISBN-10 ISBN-13

3-540-27721-8 (2nd Edition) Springer Berlin Heidelberg New York 978-3-540-27721-7 (2nd Edition) Springer Berlin Heidelberg New York 3-540-63409-6 (1st Edition) Springer Berlin Heidelberg New York 978-3-540-63409-6 (1st Edition) Springer Berlin Heidelberg New York

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com c Springer-Verlag Berlin Heidelberg 2006 Printed in The Netherlands The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: by the authors and TechBooks using a Springer LATEX macro package Cover design: design & production GmbH, Heidelberg Printed on acid-free paper

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543210

Preface to the Second Edition

Quantum mechanics is an endless source of new questions and fascinating observations. Examples can be found in fundamental physics and in applied physics, in mathematical questions as well as in the currently popular debates on the interpretation of quantum mechanics and its philosophical implications. Teaching quantum mechanics relies mostly on theoretical courses, which are illustrated by simple exercises often of a mathematical character. Reducing quantum physics to this type of problem is somewhat frustrating since very few, if any, experimental quantities are available to compare the results with. For a long time, however, from the 1950s to the 1970s, the only alternative to these basic exercises seemed to be restricted to questions originating from atomic and nuclear physics, which were transformed into exactly soluble problems and related to known higher transcendental functions. In the past ten or twenty years, things have changed radically. The development of high technologies is a good example. The one-dimensional squarewell potential used to be a rather academic exercise for beginners. The emergence of quantum dots and quantum wells in semiconductor technologies has changed things radically. Optronics and the associated developments in infrared semiconductor and laser technologies have considerably elevated the social rank of the square-well model. As a consequence, more and more emphasis is given to the physical aspects of the phenomena rather than to analytical or computational considerations. Many fundamental questions raised since the very beginnings of quantum theory have received experimental answers in recent years. A good example is the neutron interference experiments of the 1980s, which gave experimental answers to 50 year old questions related to the measurability of the phase of the wave function. Perhaps the most fundamental example is the experimental proof of the violation of Bell’s inequality, and the properties of entangled states, which have been established in decisive experiments since the late 1970s. More recently, the experiments carried out to quantitatively verify decoherence eﬀects and “Schr¨odinger-cat” situations have raised considerable

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Preface to the Second Edition

interest with respect to the foundations and the interpretation of quantum mechanics. This book consists of a series of problems concerning present-day experimental or theoretical questions on quantum mechanics. All of these problems are based on actual physical examples, even if sometimes the mathematical structure of the models under consideration is simpliﬁed intentionally in order to get hold of the physics more rapidly. ´ The problems have all been given to our students in the Ecole Polytech´ nique and in the Ecole Normale Sup´erieure in the past 15 years or so. A special ´ feature of the Ecole Polytechnique comes from a tradition which has been kept for more than two centuries, and which explains why it is necessary to devise original problems each year. The exams have a double purpose. On one hand, they are a means to test the knowledge and ability of the students. On the other hand, however, they are also taken into account as part of the entrance examinations to public oﬃce jobs in engineering, administrative and military careers. Therefore, the traditional character of stiﬀ competitive examinations and strict meritocracy forbids us to make use of problems which can be found in the existing literature. We must therefore seek them among the forefront of present research. This work, which we have done in collaboration with many colleagues, turned out to be an amazing source of discussions between us. We all actually learned very many things, by putting together our knowledge in our respective ﬁelds of interest. Compared to the ﬁrst version of this book, which was published by Springer-Verlag in 2000, we have made several modiﬁcations. First of all, we have included new themes, such as the progress in measuring neutrino oscillations, quantum boxes, the quantum thermometer etc. Secondly, it has appeared useful to include, at the beginning, a brief summary on the basics of quantum mechanics and the formalism we use. Finally, we have grouped the problems under three main themes. The ﬁrst (Part A) deals with Elementary Particles, Nuclei and Atoms, the second (Part B) with Quantum Entanglement and Measurement, and the third (Part C) with Complex Systems. We are indebted to many colleagues who either gave us driving ideas, or wrote ﬁrst drafts of some of the problems presented here. We want to pay a tribute to the memory of Gilbert Grynberg, who wrote the ﬁrst versions of “The hydrogen atom in crossed ﬁelds”, “Hidden variables and Bell’s inequalities” and “Spectroscopic measurement on a neutron beam”. We are particularly grateful to Fran¸cois Jacquet, Andr´e Roug´e and Jim Rich for illuminating discussions on “Neutrino oscillations”. Finally we thank Philippe Grangier, who actually conceived many problems among which the “Schr¨ odinger’s cat”, the “Ideal quantum measurement” and the “Quantum thermometer”, G´erald Bastard for “Quantum boxes”, Jean-No¨el Chazalviel for “Hyperﬁne structure in electron spin resonance”, Thierry Jolicoeur for “Magnetic excitons”, Bernard Equer for “Probing matter with positive muons”, Vincent Gillet for “Energy loss of ions in matter”, and Yvan Castin, Jean-Michel Courty and Do-

Preface to the Second Edition

VII

minique Delande for “Quantum reﬂection of atoms on a surface” and “Quantum motion in a periodic potential”. Palaiseau, April 2005

Jean-Louis Basdevant Jean Dalibard

Contents

Summary of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 General Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 The Particular Case of a Point-Like Particle; Wave Mechanics . 4 4 Angular Momentum and Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 5 Exactly Soluble Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 6 Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 7 Identical Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 8 Time-Evolution of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 9 Collision Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Part I Elementary Particles, Nuclei and Atoms 1

Neutrino Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Mechanism of the Oscillations; Reactor Neutrinos . . . . . . . . . . . 1.2 Oscillations of Three Species; Atmospheric Neutrinos . . . . . . . . 1.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 18 20 23 27

2

Atomic Clocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Hyperﬁne Splitting of the Ground State . . . . . . . . . . . . . . . . 2.2 The Atomic Fountain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The GPS System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 The Drift of Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . 2.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29 29 31 32 32 33

3

Neutron Interferometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Neutron Interferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The Gravitational Eﬀect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Rotating a Spin 1/2 by 360 Degrees . . . . . . . . . . . . . . . . . . . . . . . .

37 38 39 40

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3.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4

Spectroscopic Measurement on a Neutron Beam . . . . . . . . . . . 47 4.1 Ramsey Fringes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5

Analysis of a Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . 5.1 Preparation of the Neutron Beam . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Spin State of the Neutrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 The Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

Measuring the Electron Magnetic Moment Anomaly . . . . . . . 65 6.1 Spin and Momentum Precession of an Electron in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7

Decay of a Tritium Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The Energy Balance in Tritium Decay . . . . . . . . . . . . . . . . . . . . . 7.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69 69 70 71

8

The Spectrum of Positronium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Positronium Orbital States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Hyperﬁne Splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Zeeman Eﬀect in the Ground State . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Decay of Positronium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73 73 73 74 75 77

9

The Hydrogen Atom in Crossed Fields . . . . . . . . . . . . . . . . . . . . 9.1 The Hydrogen Atom in Crossed Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Pauli’s Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81

10 Energy Loss of Ions in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Energy Absorbed by One Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Energy Loss in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 88 90 94

55 55 57 57 59

82 82 83

Contents

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Part II Quantum Entanglement and Measurement 11 The EPR Problem and Bell’s Inequality . . . . . . . . . . . . . . . . . . . 99 11.1 The Electron Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 11.2 Correlations Between the Two Spins . . . . . . . . . . . . . . . . . . . . . . . 100 11.3 Correlations in the Singlet State . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 11.4 A Simple Hidden Variable Model . . . . . . . . . . . . . . . . . . . . . . . . . . 101 11.5 Bell’s Theorem and Experimental Results . . . . . . . . . . . . . . . . . . 102 11.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12 Schr¨ odinger’s Cat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12.1 The Quasi-Classical States of a Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12.2 Construction of a Schr¨ odinger-Cat State . . . . . . . . . . . . . . . . . . . . 111 12.3 Quantum Superposition Versus Statistical Mixture . . . . . . . . . . . 111 12.4 The Fragility of a Quantum Superposition . . . . . . . . . . . . . . . . . . 112 12.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 12.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13 Quantum Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.2 Correlated Pairs of Spins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.3 The Quantum Cryptography Procedure . . . . . . . . . . . . . . . . . . . . 125 13.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 14 Direct Observation of Field Quantization . . . . . . . . . . . . . . . . . . 131 14.1 Quantization of a Mode of the Electromagnetic Field . . . . . . . . 131 14.2 The Coupling of the Field with an Atom . . . . . . . . . . . . . . . . . . . 133 14.3 Interaction of the Atom with an “Empty” Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4 Interaction of an Atom with a Quasi-Classical State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 14.5 Large Numbers of Photons: Damping and Revivals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 14.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 15 Ideal Quantum Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.1 Preliminaries: a von Neumann Detector . . . . . . . . . . . . . . . . . . . . 147 15.2 Phase States of the Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . 148 15.3 The Interaction between the System and the Detector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.4 An “Ideal” Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

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15.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 16 The Quantum Eraser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 16.1 Magnetic Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 16.2 Ramsey Fringes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 16.3 Detection of the Neutron Spin State . . . . . . . . . . . . . . . . . . . . . . . 158 16.4 A Quantum Eraser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 16.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 16.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 17 A Quantum Thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.1 The Penning Trap in Classical Mechanics . . . . . . . . . . . . . . . . . . . 169 17.2 The Penning Trap in Quantum Mechanics . . . . . . . . . . . . . . . . . . 170 17.3 Coupling of the Cyclotron and Axial Motions . . . . . . . . . . . . . . . 172 17.4 A Quantum Thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Part III Complex Systems 18 Exact Results for the Three-Body Problem . . . . . . . . . . . . . . . . 185 18.1 The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 18.2 The Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 18.3 Relating the Three-Body and Two-Body Sectors . . . . . . . . . . . . . 186 18.4 The Three-Body Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . 187 18.5 From Mesons to Baryons in the Quark Model . . . . . . . . . . . . . . . 187 18.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 19 Properties of a Bose–Einstein Condensate . . . . . . . . . . . . . . . . . 193 19.1 Particle in a Harmonic Trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 19.2 Interactions Between Two Conﬁned Particles . . . . . . . . . . . . . . . . 194 19.3 Energy of a Bose–Einstein Condensate . . . . . . . . . . . . . . . . . . . . . 195 19.4 Condensates with Repulsive Interactions . . . . . . . . . . . . . . . . . . . 195 19.5 Condensates with Attractive Interactions . . . . . . . . . . . . . . . . . . . 196 19.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 19.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 20 Magnetic Excitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.1 The Molecule CsFeBr3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.2 Spin–Spin Interactions in a Chain of Molecules . . . . . . . . . . . . . . 204 20.3 Energy Levels of the Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 20.4 Vibrations of the Chain: Excitons . . . . . . . . . . . . . . . . . . . . . . . . . 206 20.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

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21 A Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 21.1 Results on the One-Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 21.2 The Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 21.3 Quantum Box in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . 218 21.4 Experimental Veriﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 21.5 Anisotropy of a Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 21.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 21.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 22 Colored Molecular Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 22.1 Hydrocarbon Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 22.2 Nitrogenous Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 22.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 22.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 23 Hyperﬁne Structure in Electron Spin Resonance . . . . . . . . . . . 237 23.1 Hyperﬁne Interaction with One Nucleus . . . . . . . . . . . . . . . . . . . . 238 23.2 Hyperﬁne Structure with Several Nuclei . . . . . . . . . . . . . . . . . . . . 238 23.3 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 23.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 24 Probing Matter with Positive Muons . . . . . . . . . . . . . . . . . . . . . . 245 24.1 Muonium in Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 24.2 Muonium in Silicon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 24.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 25 Quantum Reﬂection of Atoms from a Surface . . . . . . . . . . . . . . 255 25.1 The Hydrogen Atom–Liquid Helium Interaction . . . . . . . . . . . . . 255 25.2 Excitations on the Surface of Liquid Helium . . . . . . . . . . . . . . . . 257 25.3 Quantum Interaction Between H and Liquid He . . . . . . . . . . . . . 258 25.4 The Sticking Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 25.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 25.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 26 Laser Cooling and Trapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 26.1 Optical Bloch Equations for an Atom at Rest . . . . . . . . . . . . . . . 267 26.2 The Radiation Pressure Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 26.3 Doppler Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 26.4 The Dipole Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 26.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 26.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

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27 Bloch Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 27.1 Unitary Transformation on a Quantum System . . . . . . . . . . . . . . 277 27.2 Band Structure in a Periodic Potential . . . . . . . . . . . . . . . . . . . . . 277 27.3 The Phenomenon of Bloch Oscillations . . . . . . . . . . . . . . . . . . . . . 278 27.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 27.5 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Author Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

Summary of Quantum Mechanics

In the following pages we remind the basic deﬁnitions, notations and results of quantum mechanics.

1 Principles Hilbert Space The ﬁrst step in treating a quantum physical problem consists in identifying the appropriate Hilbert space to describe the system. A Hilbert space is a complex vector space, with a Hermitian scalar product. The vectors of the space are called kets and are noted |ψ. The scalar product of the ket |ψ1 and the ket |ψ2 is noted ψ2 |ψ1 . It is linear in |ψ1 and antilinear in |ψ2 and one has: ∗ ψ1 |ψ2 = (ψ2 |ψ1 ) . Deﬁnition of the State of a System; Pure Case The state of a physical system is completely deﬁned at any time t by a vector of the Hilbert space, normalized to 1, noted |ψ(t). Owing to the superposition principle, if |ψ1 and |ψ2 are two possible states of a given physical system, any linear combination |ψ ∝ c1 |ψ1 + c2 |ψ2 , where c1 and c2 are complex numbers, is a possible state of the system. These coeﬃcients must be chosen such that ψ|ψ = 1.

2

Summary of Quantum Mechanics

Measurement To a given physical quantity A one associates a self-adjoint (or Hermitian) operator Aˆ acting in the Hilbert space. In a measurement of the quantity A, ˆ the only possible results are the eigenvalues aα of A. Consider a system in a state |ψ. The probability P(aα ) to ﬁnd the result aα in a measurement of A is 2 P(aα ) = Pˆα |ψ , where Pˆα is the projector on the eigensubspace Eα associated to the eigenvalue aα . After a measurement of Aˆ which has given the result aα , the state of the system is proportional to Pˆα |ψ (wave packet projection or reduction). A single measurement gives information on the state of the system after the measurement has been performed. The information acquired on the state before the measurement is very “poor”, i.e. if the measurement gave the result aα , one can only infer that the state |ψ was not in the subspace orthogonal to Eα . In order to acquire accurate information on the state before measurement, one must use N independent systems, all of which are prepared in the same state |ψ (with N 1) . If we perform N1 measurements of Aˆ1 (eigenvalˆ ues p {a1,α }), N2 measurements of A2 (eigenvalues {a2,α }), and so on (with i=1 Ni = N ), we can determine the probability distribution of the ai,α , and therefore the Pˆi,α |ψ 2 . If the p operators Aˆi are well chosen, this determines unambiguously the initial state |ψ. Evolution When the system is not being measured, the evolution of its state vector is given by the Schr¨odinger equation i¯ h

d ˆ |ψ = H(t) |ψ(t) , dt

ˆ where the hermitian operator H(t) is the Hamiltonian, or energy observable, of the system at time t. If we consider an isolated system, whose Hamiltonian is time-independent, the energy eigenstates of the Hamiltonian |φn are the solution of the time independent Schr¨ odinger equation: ˆ n = En |φn . H|φ They form an orthogonal basis of the Hilbert space. This basis is particularly useful. If we decompose the initial state |ψ(0) on this basis, we can immediately write its expression at any time as:

1 Principles

|ψ(0) =

αn |φn

→

|ψ(t) =

n

3

αn e−iEn t/¯h |φn .

n

The coeﬃcients are αn = φn |ψ(0), i.e. e−iEn t/¯h |φn φn |ψ(0) . |ψ(t) = n

Complete Set of Commuting Observables (CSCO) ˆ B, ˆ . . . , X} ˆ is a CSCO if all of these operators commute A set of operators {A, and if their common eigenbasis {|α, β, . . . , ξ} is unique (up to a phase factor). In that case, after the measurement of the physical quantities {A, B, . . . , X}, the state of the system is known unambiguously. If the measurements have ˆ the state of the system is given the values α for A, β for B, . . . , ξ for X, |α, β, . . . , ξ. Entangled States Consider a quantum system S formed by two subsystems S1 and S2 . The Hilbert space in which we describe S is the tensor product of the Hilbert spaces E1 and E2 respectively associated with S1 and S2 . If we note {|αm } a basis of S1 and {|βn } a basis of S2 , a possible basis of the global system is {|αm ⊗ |βn }. Any state vector of the global system can be written as: Cm,n |αm ⊗ |βn . |Ψ = m,n

If this vector can be written as |Ψ = |α ⊗ |β, where |α and |β are vectors of E1 and E2 respectively, one calls it a factorized state. In general an arbitrary state |Ψ is not factorized: there are quantum correlations between the two subsystems, and |Ψ is called an Entangled state. Statistical Mixture and the Density Operator If we have an incomplete information on the state of the system, for instance because the measurements are incomplete, one does not know exactly its state vector. The state can be described by a density operator ρˆ whose properties are the following: • The density operator is hermitian and its trace is equal to 1. • All the eigenvalues Πn of the density operator are non-negative. The density operator can therefore be written as Πn |φn φn | , ρˆ = n

4

Summary of Quantum Mechanics

where the |φn are the eigenstates of ρˆ and the Πn can be interpreted as a probability distribution. In the case of a pure state, all eigenvalues vanish except one which is equal to 1. • The probability to ﬁnd the result aα in a measurement of the physical quantity A is given by ˆ n . Πn φn |A|φ P(aα ) = Tr Pˆα ρˆ = n

The state of the system after the measurement is ρˆ ∝ Pˆα ρˆPˆα . • As long as the system is not measured, the evolution of the density operator is given by d ˆ , ρˆ(t)] . i¯ h ρˆ(t) = [H(t) dt

2 General Results Uncertainty Relations Consider 2N physical systems which are identical and independent, and are all prepared in the same state |ψ (we assume N 1). For N of them, we measure a physical quantity A, and for the N others , we measure a physical quantity B. The rms deviations ∆a and ∆b of the two series of measurements satisfy the inequality 1 ˆ B]|ψ ˆ ∆a ∆b ≥ ψ|[A, . 2 Ehrenfest Theorem ˆ Consider a system which evolves under the action of a Hamiltonian H(t), and ˆ an observable A(t). The expectation value of this observable evolves according to the equation: d 1 ∂ Aˆ ˆ H]|ψ ˆ a = ψ|[A, + ψ| |ψ . dt i¯ h ∂t ˆ the expecIn particular, if Aˆ is time-independent and if it commutes with H, tation value a is a constant of the motion.

3 The Particular Case of a Point-Like Particle; Wave Mechanics The Wave Function For a point-like particle for which we can neglect possible internal degrees of freedom, the Hilbert space is the space of square integrable functions (written in mathematics as L2 (R3 )).

3 The Particular Case of a Point-Like Particle; Wave Mechanics

5

The state vector |ψ is represented by a wave function ψ(r). The quantity |ψ(r)|2 is the probability density to ﬁnd the particle at point r in dimensional space. Its Fourier transform ϕ(p): 1 ϕ(p) = e−ip·r/¯h ψ(r) d3 r (2π¯ h)3/2 is the probability amplitude to ﬁnd that the particle has a momentum p. Operators Among the operators associated to usual physical quantities, one ﬁnds: • The position operator rˆ ≡ (ˆ x, yˆ, zˆ), which consists in multiplying the wave function ψ(r) by r. ˆ whose action on the wave function ψ(r) is the • The momentum operator p operation −i¯ h∇. • The Hamiltonian, or energy operator, for a particle placed in a potential V (r): 2 ˆ = pˆ + V (ˆ r) H 2M

→

¯2 2 h ˆ Hψ(r) =− ∇ ψ(r) + V (r)ψ(r) , 2M

where M is the mass of the particle. Continuity of the Wave Function If the potential V is continuous, the eigenfunctions of the Hamiltonian ψα (r) are continuous and so are their derivatives. This remains true if V (r) is a step function: ψ and ψ are continuous where V (r) has discontinuities. In the case of inﬁnitely high potential steps, (for instance V (x) = +∞ for x < 0 and V (x) = 0 for x ≥ 0), ψ(x) is continuous and vanishes at the discontinuity of V (ψ(0) = 0), while its ﬁrst derivative ψ (x) is discontinuous. In one dimension, it is interesting to consider potentials which are Dirac distributions, V (x) = g δ(x). The wave function is continuous and the discontinuity of its derivative is obtained by integrating the Schr¨ odinger equation around the center of the delta function [ψ (0+ ) − ψ (0− ) = (2M g/¯h2 ) ψ(0) in our example]. Position-Momentum Uncertainty Relations Using the above general result, one ﬁnds: [ˆ x, pˆx ] = i¯ h

→

∆x ∆px ≥ ¯h/2 ,

and similar relations for the y and z components.

6

Summary of Quantum Mechanics

4 Angular Momentum and Spin Angular Momentum Observable An angular momentum observable Jˆ is a set of three operators {Jˆx , Jˆy , Jˆz } which satisfy the commutation relations h Jˆz , [Jˆx , Jˆy ] = i¯

[Jˆy , Jˆz ] = i¯ h Jˆx ,

[Jˆz , Jˆx ] = i¯h Jˆy .

ˆ = rˆ × p ˆ is an The orbital angular momentum with respect to the origin L angular momentum observable. The observable Jˆ2 = Jˆx2 + Jˆy2 + Jˆz2 commutes with all the components ˆ Ji . One can therefore ﬁnd a common eigenbasis of Jˆ2 and one of the three components Jˆi . Traditionally, one chooses i = z. Eigenvalues of the Angular Momentum The eigenvalues of Jˆ2 are of the form h ¯ 2 j(j + 1) with j integer or half integer. 2 ˆ In an eigensubspace of J corresponding to a given value of j, the eigenvalues of Jˆz are of the form hm , ¯

with m ∈ {−j, −j + 1, . . . , j − 1, j}

(2j + 1 values) .

The corresponding eigenstates are noted |α, j, m, where α represents the other quantum numbers which are necessary in order to deﬁne the states completely. The states |α, j, m are related to |α, j, m ± 1 by the operators Jˆ± = Jˆx ± iJˆy : Jˆ± |α, j, m = j(j + 1) − m(m ± 1) |α, j, m ± 1 . Orbital Angular Momentum of a Particle In the case of an orbital angular momentum, only integer values of j and m are allowed. Traditionally, one notes j = in this case. The common eigenstates ˆ z can be written in spherical coordinates as R(r) Y,m (θ, ϕ), ˆ 2 and L ψ(r) of L where the radial wave function R(r) is arbitrary and where the functions Y,m are the spherical harmonics, i.e. the harmonic functions on the sphere of radius one. The ﬁrst are: 3 1 cos θ , , Y1,0 (θ, ϕ) = Y0,0 (θ, ϕ) = √ 4π 4π 3 3 iϕ sin θ e , Y1,−1 (θ, ϕ) = sin θ e−iϕ . Y1,1 (θ, ϕ) = − 8π 8π

5 Exactly Soluble Problems

7

Spin In addition to its angular momentum, a particle can have an intrinsic angular momentum called its Spin. The spin, which is noted traditionally j = s, can take half-integer as well as integer values. The electron, the proton, the neutron are spin s = 1/2 particles, for which the projection of the intrinsic angular momentum can take either of the two values m¯h: m = ±1/2. In the basis |s = 1/2 , m = ±1/2, the operators Sˆx , Sˆy , Sˆz have the matrix representations:

h 01 ¯ h 0 −i ¯ ¯h 1 0 Sˆx = , Sˆy = , Sˆz = . 2 10 2 i 0 2 0 −1 Addition of Angular Momenta Consider a system S made of two subsystems S1 and S2 , of angular momenta ˆ 2 . The observable Jˆ = Jˆ1 + Jˆ2 is an angular momentum observable. ˆ 1 and J J In the subspace corresponding to given values j1 and j2 (of dimension (2j1 + 1) × (2j2 + 1)), the possible values for the quantum number j corresponding to the total angular momentum of the system Jˆ are: j = |j1 − j2 | , |j1 − j2 | + 1 , · · · , j1 + j2 , with, for each value of j, the 2j + 1 values of m: m = −j, −j + 1, · · · , j. For instance, adding two spins 1/2, one can obtain an angular momentum 0 (singlet state j = m = 0) and three states of angular momentum 1 (triplet states j = 1, m = 0, ±1). The relation between the factorized basis |j1 , m1 ⊗ |j2 , m2 and the total angular momentum basis |j1 , j2 ; j, m is given by the Clebsch-Gordan coeﬃcients: j,m Cj1 ,m1 ;j2 ,m2 |j1 , m1 ⊗ |j2 , m2 . |j1 , j2 ; j, m = m1 m2

5 Exactly Soluble Problems The Harmonic Oscillator For simplicity, we consider the one-dimensional problem. The harmonic potential is written V (x) = mω 2 x2 /2. The natural length and momentum scales are √ h ¯ , p0 = ¯hmω . x0 = mω ˆ =x By introducing the reduced operators X ˆ/x0 and Pˆ = pˆ/p0 , the Hamiltonian is:

8

Summary of Quantum Mechanics

¯ ω ˆ2 ˆ =h ˆ2 , H P +X 2

ˆ Pˆ ] = i . with [X,

We deﬁne the creation and annihilation operators a ˆ† and a ˆ by: 1 ˆ 1 ˆ + iPˆ , a ˆ† = √ X − iPˆ , [ˆ a, a ˆ† ] = 1 . a ˆ= √ X 2 2 One has

†

ˆ =h H ¯ω a ˆ a ˆ + 1/2 .

ˆ are (n + 1/2)¯ The eigenvalues of H hω, with n non-negative integer. These eigenvalues are non-degenerate. The corresponding eigenvectors are noted |n. We have: √ a ˆ† |n = n + 1 |n + 1 and √ a ˆ|n = n |n − 1 if n > 0 , = 0 if n = 0 . The corresponding wave functions are the Hermite functions. The ground state |n = 0 is given by: ψ0 (x) =

1 √ exp(−x2 / 2x20 ) . π 1/4 x0

Higher dimension harmonic oscillator problems are deduced directly from these results. The Coulomb Potential (bound states) We consider the motion of an electron in the electrostatic ﬁeld of the proton. We note µ the reduced mass (µ = me mp /(me + mp ) me ) and we set e2 = q 2 /(4π0 ). Since the Coulomb potential is rotation invariant, we can ﬁnd ˆ z . The bound ˆ to L ˆ 2 and to L a basis of states common to the Hamiltonian H, states are characterized by the 3 quantum numbers n, , m with: ψn,,m (r) = Rn, (r) Y,m (θ, ϕ) , where the Y,m are the spherical harmonics. The energy levels are of the form En = −

EI n2

with EI =

µe4 13.6 eV . 2¯ h2

The principal quantum number n is a positive integer and can take all integer values from 0 to n − 1. The total degeneracy (in m and ) of a given energy level is n2 (we do not take spin into account). The radial wave functions Rn, are of the form:

6 Approximation Methods

Rn, (r) = r Pn, (r) exp(−r/(na1 )) ,

with

a1 =

9

¯2 h 0.53 ˚ A. µe2

Pn, (r) is a polynomial of degree n − − 1 called a Laguerre polynomial. The length a 1 is the Bohr radius. The ground state wave function is ψ1,0,0 (r) = e−r/a1 / πa31 .

6 Approximation Methods Time-Independent Perturbations ˆ which can be written as H ˆ = We consider a time-independent Hamiltonian H ˆ 0 + λH ˆ 1 . We suppose that the eigenstates of H ˆ 0 are known: H ˆ 0 |n, r = En |n, r , H

r = 1, 2, . . . , pn

ˆ 1 is where pn is the degeneracy of En . We also suppose that the term λH suﬃciently small so that it only results in small perturbations of the spectrum ˆ 0. of H ˆ which Non-degenerate Case. In this case, pn = 1 and the eigenvalue of H coincides with En as λ → 0 is given by: ˜n = En + λ n|H ˆ 1 |n + λ2 E

|k|H ˆ 1 |n|2 + O(λ3 ) . En − E k

k=n

The corresponding eigenstate is: |ψn = |n + λ

k|H ˆ 1 |n |k + O(λ2 ) En − E k

k=n

ˆ at ﬁrst order in Degenerate Case. In order to obtain the eigenvalues of H λ, and the corresponding eigenstates, one must diagonalize the restriction of ˆ 0 associated with the eigenvalue En , i.e. ﬁnd the pn ˆ 1 to the subspace of H λH solutions of the “secular” equation: ˆ 1 |n, 1 − ∆E ˆ 1 |n, pn n, 1|λH ... n, 1|λH .. .. = 0. ˆ . . n, r|λH1 |n, r − ∆E n, p |λH ˆ ˆ ... n, pn |λH1 |n, pn − ∆E n 1 |n, 1 ˜n,r = En +∆Er , r = 1, . . . , pn . In general, The energies to ﬁrst order in λ are E the perturbation is lifted (at least partially) by the perturbation.

10

Summary of Quantum Mechanics

Variational Method for the Ground State Consider an arbitrary state |ψ normalized to 1. The expectation value of the energy in this state is greater than or equal to the ground state energy E0 : ˆ ψ|H|ψ ≥ E0 . In order to ﬁnd an upper bound to E0 , one uses a set of trial wave functions which depend on a set of parameters, and one looks for the minimum of E for these functions. This minimum always lies above E0 .

7 Identical Particles All particles in nature belong to one of the following classes: • Bosons, which have integer spin. The state vector of N identical bosons is totally symmetric with respect to the exchange of any two of these particles. • Fermions, which have half-integer spin. The state vector of N identical fermions is totally antisymmetric with respect to the exchange of any two of these particles. Consider a basis {|ni , i = 1, 2, . . .} of the one particle Hilbert space. Consider a system of N identical particles, which we number arbitrarily from 1 to N . (a) If the particles are bosons, the state vector of the system with N1 particles in the state |n1 , N2 particles in the state |n2 , etc., is: 1 1 |1 : nP (1) ; 2 : nP (2) ; . . . ; N : nP (N ) , |Ψ = √ √ N ! N1 !N2 ! · · · P where the summation is made on the N ! permutations of a set of N elements. (b) If the particles are fermions, the state corresponding to one particle in the state |n1 , another in the state |n2 , etc., is given by the Slater determinant: |1 : n1 |1 : n2 . . . |1 : nN 1 |2 : n1 |2 : n2 . . . |2 : nN |Ψ = √ . .. .. .. N ! . . . |N : n1 |N : n2 . . . |N : nN Since the state vector is antisymmetric, two fermions cannot be in the same quantum state (Pauli’s exclusion principle). The above states form a basis of the N −fermion Hilbert space.

8 Time-Evolution of Systems

11

8 Time-Evolution of Systems Rabi Oscillation ˆ0 = h ¯ ω0 |++|. We couple Consider a two-level system |±, of Hamiltonian H ˆ these two states with a Hamiltonian H1 :

hω1 −iωt ˆ1 = ¯ e H |+−| + eiωt |−+| . 2 We assume that the state of the system is |− at time t = 0. The probability to ﬁnd the system in the state |+ at time t is: P (t) =

ω12 sin2 (ΩT /2) Ω2

with Ω 2 = (ω − ω0 )2 + ω12 .

Time-Dependent Perturbation Theory ˆ 1 (t). We assume ˆ ˆ0 + H We consider a system whose Hamiltonian is H(t) =H ˆ 0 and the corresponding energies En are known. At the eigenstates |n of H ˆ 0 . To ﬁrst time t = 0, we assume that the system is in the eigenstate |i of H ˆ 1 , the probability amplitude to ﬁnd the system in another eigenstate order in H |f at time t is: 1 t i(Ef −Ei )t/¯h ˆ 1 (t )|i dt . a(t) = e f |H i¯ h 0 In the case of a time-independent perturbation H1 , the probability is: P (t) = |a(t)|2 =

2 2 1 ˆ 1 |i sin (ωt/2) , | H f (ω/2)2 h2 ¯

where we have set ¯hω = Ef − Ei . Fermi’s Golden Rule and Exponential Decay ˆ 0 . Initially, the system Consider a system with an unperturbed Hamiltonian H is in an eigenstate |i of energy Ei . We assume that this system is coupled to ˆ 0 by the time-independent perturbation a continuum {|f } of eigenstates of H Vˆ . For simplicity, we assume that the matrix elements f |Vˆ |i only depend on the energies Ef of the states |f . To lowest order in Vˆ , this coupling results in a ﬁnite lifetime τ of the state |i: the probability to ﬁnd the system in the state |i at time t > 0 is e−t/τ with: 2π 1 = |f |Vˆ |i|2 ρ(Ei ) . τ h ¯ The matrix element f |Vˆ |i is evaluated for a state |f of energy Ef = Ei . The function ρ(E) is the density of ﬁnal states. For non relativistic particles

12

Summary of Quantum Mechanics

(E = p2 /2m) or ultra-relativistic particles (E = cp, for instance photons), its values are respectively: √ mL3 2mE L3 E 2 ρ (E) = . ρnon rel. (E) = ultra rel. 2π 2 ¯ h3 2π 2 ¯h3 c3 When the spin degree of freedom of the particle comes into play, this density of state must be multiplied by the number of possible spin states 2s + 1, where s is the spin of the particle. The quantity L3 represents the normalization volume (and cancels identically with the normalization factors of the states |i and |f ). Consider an atomic transition treated as a two-level system, an excited state |e and a ground state |g, separated by an energy h ¯ ω and coupled via an electric dipole interaction. The lifetime τ of the excited state due to this spontaneous emission is given by: ω3 1 ˆ 2 e| D|g = , τ 3π0 ¯ hc3 ˆ is the electric dipole operator. where D

9 Collision Processes Born Approximation We consider an elastic collision process of a non-relativistic particle of mass m with a ﬁxed potential V (r). To second order in V , the elastic scattering cross-section for an incident particle in the initial momentum state p and the ﬁnal momentum state p is given by: dσ = dΩ

m 2π¯ h2

2

|V˜ (p − p )|2 ,

with V˜ (q) =

eiq·r/¯h V (r) d3 r .

Example: the Yukawa potential. We consider V (r) = g

¯ c −r/a h e , r

which gives, writing p = h ¯ k: dσ = dΩ

2mgca2 h ¯

2

1 1+

4a2 k 2

2 sin2 (θ/2)

(Born) ,

where θ is the scattering angle between p and p . The total cross-section is then:

2 4πa2 2 mgca (Born) . σ(k) = h ¯ 1 + 4k 2 a2

9 Collision Processes

13

In the case where the range a of the potential tends to inﬁnity, we recover the Coulomb cross section:

2 g¯ hc 1 dσ = (exact) , dΩ 4E sin4 (θ/2) where E = p2 /(2m). Scattering by a Bound State We consider a particle a of mass m undergoing an elastic scattering on a system composed of n particles b1 , . . . , bn . These n particles form a bound state whose wave function is ψ0 (r 1 , . . . , r n ). In Born approximation, the cross section is

2 m dσ V˜j (q) Fj (q) . = |V(p − p )|2 with V(q) = 2 dΩ 2π¯ h j The potential Vj represents the interaction between particles a and bj . The form factor Fj is deﬁned by: Fj (q) = eiq·rj /¯h |ψ0 (r 1 , ..., r j , ..., r n )|2 d3 r1 . . . d3 rj . . . d3 rn . In general, interference eﬀects can be observed between the various q contributing to the sum which deﬁnes V(q). In the case of a charge distribution, V˜ is the Rutherford amplitude, and the form factor F is the Fourier transform of the charge density. General Scattering Theory In order to study the general problem of the scattering of a particle of mass m by a potential V (r), it is useful to determine the positive energy E = ˆ = pˆ2 /(2m) + V (r) whose asymptotic form is h2 k 2 /(2m) eigenstates of H ¯ ψk (r)

∼

|r|→∞

eik·r + f (k, u, u )

eikr . r

This corresponds to the superposition of an incident plane wave eik·r and a scattered wave. Such a state is called a stationary scattering state. The scattering amplitude f depends on the energy, on the incident direction u = k/k, and on the ﬁnal direction u = r/r. The diﬀerential cross section is given by: dσ = |f (k, u, u )|2 . dΩ The scattering amplitude is given by the implicit equation m e−ik ·r V (r ) ψk (r ) d3 r with k = ku . f (k, u, u ) = − 2 2π¯ h We recover Born’s approximation by choosing ψk (r ) eik·r .

14

Summary of Quantum Mechanics

Low Energy Scattering When the wavelength of the incident particle λ ∼ k −1 is large compared to the range of the potential, the amplitude f does not depend on u and u (at least if the potential decreases faster than r−3 at inﬁnity). The scattering is isotropic. The limit as = − limk→0 f (k) is called the scattering length.

Part I

Elementary Particles, Nuclei and Atoms

2 Atomic Clocks

We are interested in the ground state of the external electron of an alkali atom (rubidium, cesium,...). The atomic nucleus has a spin sn (sn = 3/2 for 87 Rb, sn = 7/2 for 133 Cs), which carries a magnetic moment µn . As in the case of atomic hydrogen, the ground state is split by the hyperﬁne interaction between the electron magnetic moment and the nuclear magnetic moment µn . This splitting of the ground state is used to devise atomic clocks of high accuracy, which have numerous applications such as ﬂight control in aircrafts, the G.P.S. system, the measurement of physical constants etc. In all the chapter, we shall neglect the eﬀects due to internal core electrons.

2.1 The Hyperﬁne Splitting of the Ground State 2.1.1. Give the degeneracy of the ground state if one neglects the magnetic interaction between the nucleus and the external electron. We note |me ; mn = |electron: se = 1/2, me ⊗ |nucleus: sn , mn a basis of the total spin states (external electron + nucleus). 2.1.2. We now take into account the interaction between the electron magnetic moment µe and the nuclear magnetic moment µn . As in the hydrogen atom, one can write the corresponding Hamiltonian (restricted to the spin subspace) as: ˆn , ˆe · S ˆ = A S H h2 ¯ ˆ e and S ˆ n are the spin operators where A is a characteristic energy, and where S of the electron and the nucleus, respectively. We want to ﬁnd the eigenvalues of this Hamiltonian. We introduce the operators Sˆe,± = Sˆe,x ± iSˆe,y and Sˆn,± = Sˆn,x ± iSˆn,y .

30

2 Atomic Clocks

(a) Show that ˆ = A Sˆe,+ Sˆn,− + Sˆe,− Sˆn,+ + 2Sˆe,z Sˆn,z . H 2 2¯ h (b) Show that the two states |me = 1/2; mn = sn

and

|me = −1/2; mn = −sn

ˆ and give the corresponding eigenvalues. are eigenstates of H, ˆ on the state |me = 1/2; mn with mn = sn ? (c) What is the action of H ˆ What is the action of H on the state |me = −1/2; mn with mn = −sn ? ˆ can be calculated (d) Deduce from these results that the eigenvalues of H by diagonalizing 2 × 2 matrices of the type:

A mn sn (sn + 1) − mn (mn + 1) . sn (sn + 1) − mn (mn + 1) −(mn + 1) 2 ˆ splits the ground state in two substates of energies E1 = 2.1.3. Show that H E0 + Asn /2 and E2 = E0 − A(1 + sn )/2. Recover the particular case of the hydrogen atom. 2.1.4. What are the degeneracies of the two sublevels E1 and E2 ? 2.1.5. Show that the states of energies E1 and E2 are eigenstates of the 2 ˆe + S ˆ n . Give the corresponding value s of ˆ2 = S square of the total spin S the spin.

H=1 m

Electromagnetic cavity Cold atoms Fig. 2.1. Sketch of the principle of an atomic clock with an atomic fountain, using laser-cooled atoms

2.2 The Atomic Fountain

31

2.2 The Atomic Fountain The atoms are initially prepared in the energy state E1 , and are sent upwards (Fig. 2.1). When they go up and down they cross a cavity where an electromagnetic wave of frequency ω is injected. This frequency is close to h. At the end of the descent, one detects the number of ω0 = (E1 − E2 )/¯ atoms which have ﬂipped from the E1 level to the E2 level. In all what follows, the motion of the atoms in space (free fall) is treated classically. It is only the evolution of their internal state which is treated quantum-mechanically. In order to simplify things, we consider only one atom in the sub-level of energy E1 . This state (noted |1) is coupled by the electromagnetic wave to only one state (noted |2) of the sublevel of energy E2 . By convention, we ﬁx ¯ ω0 /2, E2 = −¯ hω0 /2. We the origin of energies at (E1 + E2 )/2, i.e. E1 = h assume that the time to cross the cavity is very brief and that this crossing results in an evolution of the state vector of the form: |ψ(t) = α|1 + β|2 −→ |ψ(t + ) = α |1 + β |2 ,

1 α α 1 −ie−iωt √ with: . = iωt β 1 β 2 −ie 2.2.1. The initial state of the atom is |ψ(0) = |1. We consider a single round-trip of duration T , during which the atom crosses the cavity between t = 0 and t = , then evolves freely during a time T − 2, and crosses the cavity a second time between T − and T . Taking the limit → 0, show that the state of the atom after this round-trip is given by: |ψ(T ) = i e−iωT /2 sin((ω − ω0 )T /2) |1 − i eiωT /2 cos((ω − ω0 )T /2) |2 (2.1) 2.2.2. Give the probability P (ω) to ﬁnd an atom in the state |2 at time T . Determine the half-width ∆ω of P (ω) at the resonance ω = ω0 . What is the values of ∆ω for a 1 meter high fountain? We recall the acceleration of gravity g = 9.81 ms−2 . 2.2.3. We send a pulse of N atoms (N 1). After the round-trip, each atom is in the state given by (2.1). We measure separately the numbers of atoms in the states |1 and |2, which we note N1 and N2 (with N1 + N2 = N ). What is the statistical distribution of the random variables N1 and N2 ? Give their mean values and their r.m.s. deviations ∆Ni . Set φ = (ω − ω0 )T /2 and express the results in terms of cos φ, sin φ and N . 2.2.4. The departure from resonance |ω − ω0 | is characterized by the value of cos((ω − ω0 )T ) = N2 − N1 /N . Justify this formula. Evaluate the uncertainty ∆|ω − ω0 | introduced by the random nature of the variable N2 − N1 . Show that this uncertainty depends on N , but not on φ. 2.2.5. In Fig. 2.2 we have represented the precision of an atomic clock as a function of the number N of atoms per pulse. Does this variation with N agree with the previous results?

32

2 Atomic Clocks

Relative precision 10-13

N 10-14 104

105

106

Fig. 2.2. Relative accuracy ∆ω/ω of a fountain atomic clock as a function of the number of atoms N sent in each pulse

2.3 The GPS System The GPS system uses 24 satellites orbiting around the Earth at 20 000 km. Each of them contains an atomic clock. Each satellite sends, at equal spaced time intervals, an electromagnetic signal composed of a “click” from a clock and the indication of its position. A reception device on Earth, which does not have an atomic clock, detects the signals coming from several satellites. With its own (quartz) clock, it compares the times at which diﬀerent “clicks” arrive. 2.3.1. What is the minimum number of satellites that one must see at a given time in order to be able to position oneself in latitude, in longitude, and in altitude on the surface of the Earth? 2.3.2. We assume that the relative accuracy of each clock is ∆ω/ω = 10−13 and that the clocks are synchronized every 24 hours. What is the order of magnitude of the accuracy of the positioning just before the clocks undergo a new synchronization?

2.4 The Drift of Fundamental Constants Some cosmological models predict a (small) variation in time of the ﬁne struchc) ∼ 1/137. In order to test such an assumption, one ture constant α = e2 /(¯ can compare two atomic clocks, one using rubidium (Z = 37) atoms, the other cesium (Z = 55) atoms. In fact, one can show that the hyperﬁne splitting of an alkali atom varies approximately as:

11 2 2 ¯ ω0 ∝ α 1 + (αZ) for (αZ)2 1 . E1 − E2 = h 6

2.5 Solutions

33

By comparing a rubidium and a cesium clock at a one year interval, no sig(Cs) (Rb) was observed. More precisely, niﬁcant variation of the ratio R = ω0 /ω0 the relative variation |δR|/R is smaller than the experimental uncertainty, estimated to be 3 × 10−15 . What upper bound can one set on the relative variation rate |α/α|? ˙

2.5 Solutions Section 2.1: Hyperﬁne Splitting of the Ground State 2.1.1. The Hilbert space of the ground state is the tensor product of the electron spin space and the nucleus spin space. Its dimension d is therefore the product of their dimensions, i.e. d = 2 × (2sn + 1). 2.1.2. Energy levels of the hyperﬁne Hamiltonian. (a) Making use of 1 ˆ i ˆ Se,+ + Sˆe,− , Sˆe,y = Se,− − Sˆe,+ , Sˆe,x = 2 2 and a similar relation for Sˆn,x and Sˆn,y , one obtains the wanted result. (b) The action of Sˆe,+ Sˆn,− and Sˆe,− Sˆn,+ on |me = 1/2; mn = sn gives the null vector. The same holds for |me = −1/2; mn = −sn . Therefore, only the term contributes Sˆe,z Sˆn,z and one ﬁnds: ˆ |me = 1/2; mn = sn = Asn |me = 1/2; mn = sn H 2 ˆ |me = −1/2; mn = −sn = Asn |me = −1/2; mn = −sn . H 2 (c) We ﬁnd: Amn ˆ H|1/2; mn = |1/2; mn 2 A + sn (sn + 1) − mn (mn + 1) | − 1/2; mn + 1 2 ˆ − 1/2; mn = − Amn | − 1/2; mn H| 2 A + sn (sn + 1) − mn (mn − 1) |1/2; mn − 1 . 2 (d) >From the previous question, one concludes that the 2-dimensional subspaces Emn generated by |1/2; mn and | − 1/2; mn + 1 are globally stable ˆ The determination of the eigenvalues of H ˆ therefore under the action of H. consists in diagonalizing the series of 2 × 2 matrices corresponding to its reˆ striction to these subspaces. The matrix corresponding to the restriction of H to the subspace Emn is the same as given in the text.

34

2 Atomic Clocks

2.1.3. The eigenvalues given in the text are actually independent of mn . They are Asn /2 and −A(1 + sn )/2. In the case sn = 1/2 (hydrogen atom), these two eigenvalues are A/4 and −3A/4. 2.1.4. There are 2sn 2 × 2 matrices to be diagonalized, each of which gives a vector associated to Asn /2 and a vector associated to −A(1 + sn )/2. In addition we have found two independent eigenvectors, |1/2, sn and | − 1/2, −sn , associated to the eigenvalue Asn /2. We therefore obtain: Asn /2

degenerated 2sn + 2 times

−A(1 + sn /2)

degenerated 2sn times

We do recover the dimension 2(2sn + 1) of the total spin space of the ground state. 2.1.5. The square of the total spin is: 2 ˆ2 + S ˆ 2 + 2S ˆe · S ˆn = S ˆ2 + S ˆ 2 + 2¯h H ˆ2 = S ˆ . S e n e n A

ˆ 2 are proportional to the identity and are respectively: ˆ 2 and S The operators S e n h2 ˆ 2 = 3¯ S e 4

ˆ2 = h ¯ 2 sn (sn + 1) . S n 2

ˆ . More precisely, an eigenˆ is therefore an eigenstate of S An eigenstate of H ˆ 2 with eigenvalue ˆ state of H with eigenvalue Asn /2 is an eigenstate of S h2 (sn + 1/2)(sn + 3/2), corresponding to a total spin s = sn + 1/2. An eigen¯ ˆ 2 with eigenvalue ˆ with eigenvalue −A(1+sn )/2 is an eigenstate of S state of H h2 (sn − 1/2)(sn + 1/2), i.e. a total spin s = sn − 1/2. ¯ Section 2.2: The Atomic Fountain 2.2.1. In the limit → 0, the ﬁnal state vector of the atom is simply the matrix product: −iω T /2

1 e 0 0 α 1 −ie−iωT × = β 1 0 eiω0 T /2 2 −ieiωT

1 −i 1 × , −i 1 0 which corresponds to crossing the cavity, at time t = 0, then to a free evolution between t = 0 and t = T , then a second crossing of the cavity at time t = T . We therefore obtain the state vector of the text. 2.2.2. One ﬁnds P (ω) = |β |2 = cos2 ((ω − ω0 )T /2). This probability is equal π/(2T ). to 1 if one sits exactly at the resonance (ω = ω0 ). It is 1/2 if ω = ω0 ± For a round-trip free fall motion of height H = 1 m, we have T = 2 2H/g, i.e. T = 0.9 s and ∆ω = 1.7 s−1 .

2.5 Solutions

35

2.2.3. The detection of each atom gives the result E1 with a probability sin2 φ and E2 with a probability cos2 φ. Since the atoms are assumed to be independent, the distributions of the random variables N1 and N2 are binomial laws. We therefore have: √ N2 = N cos2 φ ∆N1 = ∆N2 = N | cos φ sin φ| . N1 = N sin2 φ 2.2.4. We do obtain N2 −N1 /N = cos 2φ = cos((ω −ω0 )T ). The ﬂuctuation on the variable N2 − N1 induces a ﬂuctuation on the determination of ω − ω0 . The two ﬂuctuations are related by: ∆(N2 − N1 ) = 2 |sin(2φ)| ∆φ . N √ √ Since ∆(N2 − N1 ) = 2 ∆N2 = N |sin 2φ|, we deduce ∆φ = 1/(2 N ), or equivalently: 1 √ . ∆|ω − ω0 | = 2T N The longer the time T and the larger N are, the better the accuracy. 2.2.5. We notice on Fig. 2.2 that the accuracy of the clock improves like N −1/2 , as N increases. For N = 106 and T = 0.9 s, the above formula gives 5.6 × 10−4 s. The hyperﬁne frequency of cesium is ω0 = 2π × 9.2 GHz, which corresponds to ∆ω/ω ∼ 10−14 . Section 2.3: The GPS System 2.3.1. One must see at least 4 satellites. With two of them, the diﬀerence between the two reception times t1 and t2 of the signals localize the observer on a surface (for instance on a plane at equal distances of the two satellites if t1 = t2 ); three satellites localize the observer on a line, and the fourth one determines the position of the observer unambiguously (provided of course that one assumes the observer is not deep inside the Earth or on a far lying orbit). 2.3.2. Suppose a satellite sends a signal at time t0 . This signal is received by an observer at a distance D at time t1 = t0 + D/c. If the clock of the satellite has drifted, the signal is not sent at time t0 , but at a slightly diﬀerent time t0 . The observer whom we assume has a correct time reference from another satellite, interprets the time t1 − t0 as a distance D = c(t1 − t0 ), he therefore makes an error c(t0 −t0 ) on his position. For a clock of relative accuracy 10−13 , the typical drift after 24 hours (=86 000 seconds) is 86 000 × 10−13 s, i.e. an error on the position of 2.5 meters. Note that the atomic clocks boarding the GPS satellites are noticeably less accurate than the fountain cold atom clocks in ground laboratories.

36

2 Atomic Clocks

Section 2.4: The Drift of Fundamental Constants Using the expression given in the text for the dependence on α of the frequencies ωCs and ωRb , we ﬁnd that a variation of the ratio R would be related to the variation of α by: 2 2 1 dα 11α2 ZCs − ZRb 1 dR = . R dt α dt 3 (1 + 11(αZRb )2 /6) (1 + 11(αZCs )2 /6) The quantity inside the brackets is 0.22, which leads to an upper bound of α/α ˙ of 1.4 × 10−14 per year, i.e. 4.3 × 10−22 per second. If we extrapolate this variation time to a time of the order of the age of the universe, this corresponds to a variation of 10−4 . Such an eﬀect should be detectable, in principle, by spectroscopic measurements on very far objects. Remark: a more precise determination of the α dependence of ωCs , for which the approximation Zα 1 is not very good, gives for the quantity inside the bracket a value of 0.45. Section 2.5: References The experimental data on the stability of a cold atom clock have been taken from the paper from the group of A. Clairon and C. Salomon, at Observatoire de Paris: G. Santarelli et al., Phys. Rev. Lett. 82, 4619 (1999). Concerning the drift of fundamental constants, see J. D. Prestage, R. L. Tjoelker, and L. Maleki, Phys. Rev. Lett. 74, 3511 (1995); H. Marion et al., Phys. Rev. Lett. 90, 150801 (2003); M. Fischer et al., Phys. Rev. Lett. 92, 230802 (2004).

3 Neutron Interferometry

In the late 1970s, Overhauser and his collaborators performed several neutron interference experiments which are of fundamental importance in quantum mechanics, and which settled debates which had started in the 1930s. We study in this chapter two of these experiments, aiming to measure the inﬂuence on the interference pattern (i) of the gravitational ﬁeld and (ii) of a 2π rotation of the neutron wave function. We consider here an interferometer made of three parallel, equally spaced crystalline silicon strips, as shown in Fig. 3.1. The incident neutron beam is assumed to be monochromatic.

Fig. 3.1. The neutron interferometer: The three “ears” are cut in a silicon monocrystal; C2 and C3 are neutron counters

For a particular value of the angle of incidence θ, called the Bragg angle, a plane wave ψinc = ei(p·r−Et)/¯h , where E is the energy of the neutrons and p their momentum, is split by the crystal into two outgoing waves which are symmetric with respect to the perpendicular direction to the crystal, as shown in Fig. 3.2.

38

3 Neutron Interferometry

Fig. 3.2. Splitting of an incident plane wave satisfying the Bragg condition

The transmitted wave and the reﬂected wave have complex amplitudes which can be written respectively as α = cos χ and β = i sin χ, where the angle χ is real: ψI = αei(p·r−Et)/¯h

ψII = βei(p ·r−Et)/¯h ,

(3.1)

where |p| = |p | since the neutrons scatter elastically on the nuclei of the crystal. The transmission and reﬂection coeﬃcients are T = |α|2 and R = |β|2 , with of course T + R = 1. In the interferometer shown in Fig. 3.1, the incident neutron beam is horizontal. It is split by the interferometer into a variety of beams, two of which recombine and interfere at point D. The detectors C2 and C3 count the outgoing neutron ﬂuxes. The neutron beam velocity corresponds to a de Broglie wavelength λ = 1.445 ˚ A. We recall the value of neutron mass M = 1.675×10−27 kg. The neutron beam actually corresponds to wave functions which are quasimonochromatic and which have a ﬁnite extension in the transverse directions. In order to simplify the writing of the equations, we only deal with pure monochromatic plane waves, as in (3.1).

3.1 Neutron Interferences 3.1.1. The measured neutron ﬂuxes are proportional to the intensities of the waves that reach the counters. Deﬁning the intensity of the incoming beam to be 1 (the units are arbitrary), write the amplitudes A2 and A3 of the wave functions which reach the counters C2 and C3 , in terms of α and β (it is not necessary to write the propagation terms ei(p·r−Et)/¯h ). Calculate the measured intensities I2 and I3 in terms of the coeﬃcients T and R. 3.1.2. Suppose that we create a phase shift δ of the wave propagating along AC, i.e. in C the wave function is multiplied by eiδ . (a) Calculate the new amplitudes A2 and A3 in terms of α, β and δ. (b) Show that the new measured intensities I2 and I3 are of the form I2 = µ − ν(1 + cos δ)

I3 = ν(1 + cos δ)

and express µ and ν in terms of T and R. (c) Comment on the result for the sum I2 + I3 .

3.2 The Gravitational Eﬀect

39

3.2 The Gravitational Eﬀect The phase diﬀerence δ between the beams ACD and ABD is created by rotating the interferometer by an angle φ around the direction of incidence. This creates a diﬀerence in the altitudes of BD and AC, which both remain horizontal, as shown in Fig. 3.3. The diﬀerence in the gravitational potential energies induces a gravitational phase diﬀerence. 3.2.1. Let d be the distance between the silicon strips, whose thickness is neglected here. Show that the side L of the lozenge ABCD and its height H, shown in Fig. 3.3, are related to d and to the Bragg angle θ by L = d/ cos θ and H = 2d sin θ. Experimentally the values of d and θ are d = 3.6 cm and θ = 22.1◦ .

Fig. 3.3. Turning the interferometer around the incident direction, in order to observe gravitational eﬀects

3.2.2. For an angle φ, we deﬁne the gravitational potential V to be V = 0 along AC and V = V0 along BD. (a) Calculate the diﬀerence ∆p of the neutron momenta in the beams AC and BD (use the approximation ∆p p). Express the result in terms of the momentum p along AC, the height H, sin φ, M , and the acceleration of gravity g. √ (b) Evaluate numerically the velocity 2gH. How good is the approximation ∆p p? 3.2.3. Evaluate the phase diﬀerence δ between the paths ABD and ACD. One can proceed in two steps: (a) Compare the path diﬀerence between the segments AB and CD. (b) Compare the path diﬀerence between the segments BD and AC. 3.2.4. The variation with φ of the experimentally measured intensity I2 in the counter C2 is represented in Fig. 3.4. (The data does not display a minimum exactly at φ = 0 because of calibration diﬃculties.) Deduce from these data the value of the acceleration due to gravity g.

40

3 Neutron Interferometry

Fig. 3.4. Measured neutron intensity in counter C2 as the angle φ is varied

3.3 Rotating a Spin 1/2 by 360 Degrees The plane of the setup is now horizontal. The phase diﬀerence arises by placing along AC a magnet of length l which produces a constant uniform magnetic ﬁeld B 0 directed along the z axis, as shown in Fig. 3.5.

Fig. 3.5. Experimental setup for observing the neutron spin Larmor precession

The neutrons are spin-1/2 particles, and have an intrinsic magnetic moˆ is the neutron spin operator, and the σ ˆ = µ0 σ ˆ where S ˆ = γn S ˆi ment µ (i = x, y, z) are the usual 2 × 2 Pauli matrices. The axes are represented in Fig. 3.5: the beam is along the y axis, the z axis is in the ABCD plane, and the x axis is perpendicular to this plane. We assume that the spin variables and the space variables are uncorrelated, i.e. at any point in space the wave function factorizes as

a+ (t) ψ+ (r, t) i(p·r−Et)/¯ h =e . ψ− (r, t) a− (t) We neglect any transient eﬀect due to the entrance and the exit of the ﬁeld zone.

3.3 Rotating a Spin 1/2 by 360 Degrees

41

The incident neutrons are prepared in the spin state

1 1 | + x = √ , 2 1 which is the eigenstate of µ ˆx with eigenvalue +µ0 . The spin state is not modiﬁed when the neutrons cross the crystal strips. 3.3.1. (a) Write the magnetic interaction Hamiltonian of the spin with the magnetic ﬁeld. (b) What is the time evolution of the spin state of a neutron in the magnet? h, calculate the three components of the expecta(c) Setting ω = −2µ0 B0 /¯ ˆ in this state, and describe the time evolution of µ ˆ in the tion value µ magnet. 3.3.2. When the neutron leaves the magnet, what is the probability Px (+µ0 ) of ﬁnding µx = +µ0 when measuring the x component of the neutron magnetic moment? For simplicity, one can set T = M lλ/(2π¯h) and express the result in terms of the angle δ = ωT /2. 3.3.3. For which values bn = nb1 (n integer) of the ﬁeld B0 is this probability equal to 1? To what motion of the average magnetic moment do these values bn correspond? A. Calculate b1 with µ0 = −9.65 × 10−27 J T−1 , l = 2.8 cm, λ = 1.445 ˚ 3.3.4. Write the state of the neutrons when they arrive on C2 and C3 (note p2 and p3 the respective momenta). 3.3.5. The counters C2 and C3 measure the neutron ﬂuxes I2 and I3 . They are not sensitive to spin variables. Express the diﬀerence of intensities I2 − I3 in terms of δ and of the coeﬃcients T and R. 3.3.6. The experimental measurement of I2 − I3 as a function of the applied ﬁeld B0 is given in Fig. 3.6. A numerical ﬁt of the curve shows that the distance between two maxima is ∆B = (64 ± 2) × 10−4 T.

Fig. 3.6. Diﬀerence of counting rates (I2 − I3 ) as a function of the applied ﬁeld

42

3 Neutron Interferometry

Comparing the values bn of question 3.3 with this experimental result, and recalling the result of a measurement of µx for these values, explain why this proves that the state vector of a spin-1/2 particle changes sign under a rotation by an odd multiple of 2π.

3.4 Solutions Section 3.1: Neutron Interferences 3.1.1. The beams ABDC2 and ACDC2 interfere. Omitting the propagation factors, one has, at C2 an amplitude A2 = α2 β + β 3 = β(α2 + β 2 ) . Similarly, for ABDC3 and ACDC3 , A3 = 2αβ 2 . The intensities at the two counters are I2 = R − 4R2 T

I3 = 4R2 T .

3.1.2. When there is a phase shift δ in C, the above expressions get modiﬁed as follows: A2 = α2 βeiδ + β 3 = β(α2 eiδ + β 2 ) ,

A3 = αβ 2 (1 + eiδ ) .

The intensities become I2 = R − 2R2 T (1 + cos δ)

I3 = 2R2 T (1 + cos δ) .

The fact that I2 + I3 does not depend on the phase shift δ is a consequence of the conservation of the total number of particles arriving at D. Section 3.2: The Gravitational Eﬀect 3.2.1. This results from elementary trigonometry. (a) Since there is no recoil energy of the silicon atoms to be taken care of, the neutron total energy (kinetic+potential) is a constant of the motion in all the 2 process. This energy is given by EAC = p2 /2M and EBD = (p − ∆p) /2M + M gH sin φ, hence ∆p M 2 gH sin φ/p . √ (b) The velocity 2gH is of the order of 0.5 m/s, and the neutron velocity is v = h/M λ 2700 m/s. The change in velocity ∆v is therefore very small: ∆v = gH/v 2 × 10−4 m/s for φ = π/2.

3.4 Solutions

43

3.2.2. (a) The gravitational potential varies in exactly the same way along AB and CD. The neutron state in both cases is a plane wave with momentum p = h/λ just before A or C. The same Schr¨odinger equation is used to determine the wave function at the end of the segments. This implies that the phases accumulated along the two segments AB and CD are equal. (b) When comparing the segments AC and BD, the previous reasoning does not apply, since the initial state of the neutron is not the same for the two segments. The initial state is exp(ipz/¯h) for AC, and exp[i(p − ∆p)z/¯h] for BD. After travelling over a distance L = AC = BD, the phase diﬀerence between the two paths is δ=

M 2 gλd2 ∆p L = tan θ sin φ . h ¯ π¯ h2

3.2.3. >From the previous result, one has δ2 − δ1 = A g (sin φ2 − sin φ1 ), h2 ). Therefore, where A = M 2 λd2 tan θ/(π¯ g=

δ2 − δ1 . A (sin φ2 − sin φ1 )

There are 9 oscillations, i.e. (δ2 − δ1 ) = 18π, between φ1 = −32◦ and φ2 = +24◦ , which gives g 9.8 ms−2 . The relative precision of the experiment was actually of the order of 10−3 . Section 3.3: Rotating a Spin 1/2 by 360 Degrees 3.3.1. Since B is along the z axis, the magnetic Hamiltonian is:

hω 1 0 ¯ ˆ HM = −µ · B 0 = . 0 −1 2 At time t, the spin state is 1 |Σ(t) = √ 2

e−iωt/2 e+iωt/2

.

d By a direct calculation of µ or by using Ehrenfest theorem ( dt µ = 1 ˆ ˆ [ µ, H]), we obtain: i¯ h

dµx = ωµy dt

dµy = −ωµx dt

dµz =0. dt

Initially µx = µ0 and µy = µz = 0; therefore, µ = µ0 (cos ωt ux + sin ωt uy ) .

44

3 Neutron Interferometry

3.3.2. When the neutrons leave the ﬁeld zone, the probability of ﬁnding µx = +µ0 is Px (+µ0 ) = |+x|Σ(T )|2 = cos2

ωT = cos2 δ 2

with T = l/v = lM λ/h. 3.3.3. The above probability is equal to 1 if δ = nπ (ωT = 2nπ), or B0 = nb1 with h2 2π 2 ¯ = 34.5 × 10−4 T . b1 = µ0 M lλ For δ = nπ the magnetic moment has rotated by 2nπ around the z axis by Larmor precession. 3.3.4. The formulas are similar to those found in question 1.2. The phase of the upper component of the spinor written in the {|+z , |−z } basis, is shifted by +δ, that of the lower component by −δ:

2 β β + α2 eiδ Amplitude at the counter C2 : ei(p2 ·r−Et)/¯h √ 2 2 −iδ 2 β +α e Amplitude at the counter C3 :

αβ 2 ei(p3 ·r−Et)/¯h √ 2

1 + eiδ 1 + e−iδ

3.3.5. Since the measuring apparatus is insensitive to spin variables, we must add the probabilities corresponding to Sz = ±1, each of which is the modulus squared of a sum of amplitudes. Altogether, we obtain the following intensities of the total neutron ﬂux in the two counters: I2 = R − 2R2 T (1 + cos δ) ,

I3 = 2R2 T (1 + cos δ)

and I2 − I3 = R − 4R2 T (1 + cos δ) . 3.3.6. There will be a minimum of I2 − I3 each time cos δ = +1, i.e. δ = 2nπ. This corresponds to a constructive interference in channel 3. On the other hand, there appears a maximum if cos δ = −1, i.e. δ = (2n + 1)π, and this corresponds to a destructive interference in channel 3 (I3 = 0). If δ = nπ, whatever the integer n, one is sure to ﬁnd the neutrons in the same spin state as in the initial beam. However, the interference pattern depends on the parity of n. The experimental result ∆B = (64 ± 2) × 10−4 T conﬁrms that if the spin has rotated by 4nπ, one recovers a constructive interference in channel 3 as in the absence of rotation, while if it has rotated by (4n + 2)π, the interference in C3 is destructive. The probability amplitude for the path ACD has changed sign in this latter case, although a spin measurement in this path after the magnet will give exactly the same result as on the incoming beam.

3.4 Solutions

45

References A.W. Overhauser, A.R. Collela, and S.A. Werner, Phys. Rev. Lett., 33, 1237 (1974); 34, 1472 (1975); 35, 1053 (1975). See also D. Greenberger and A.W. Overhauser, Scientiﬁc American, May 1980.

4 Spectroscopic Measurement on a Neutron Beam

We present here a very precise method for spectroscopic measurements, due to Norman Ramsey. The method, using atomic or molecular beams, can be applied to a very large class of problems. We shall analyse it in the speciﬁc case of a neutron beam, where it can be used to determine the neutron magnetic moment with high accuracy, by measuring the Larmor precession frequency in a magnetic ﬁeld B 0 . A beam of neutrons is prepared with velocity v along the x axis. The beam is placed in a constant uniform magnetic ﬁeld B 0 directed along the z axis. We write |+ and |− for the eigenstates of the z projection Sˆz of the neutron ˆ µ ˆ = γ S, ˆ being the spin, and γ for the gyromagnetic ratio of the neutron: µ ˆ neutron magnetic moment operator, and S its spin. The neutrons are initially in the state |−. When they approach the origin, they cross a zone where an oscillating ﬁeld B 1 (t) is applied in the (x, y) plane. The components of B1 are B1x = B1 e−r/a cos ω(t − z/c) B1y = B1 e−r/a sin ω(t − z/c) B1z = 0 ,

(4.1)

where r = x2 + y 2 . We assume that B1 is constant (strictly speaking it should vary in order to satisfy ∇ · B = 0) and that B1 B0 . In all parts of the chapter, the neutron motion in space is treated classically as a linear uniform motion. We are only interested in the quantum evolution of the spin state.

4.1 Ramsey Fringes 4.1.1. Consider a neutron whose motion in space is x = vt, y = 0, z = 0. ˆ What is the Hamiltonian H(t) describing the coupling of the neutron magnetic moment with the ﬁelds B 0 and B 1 ?

48

4 Spectroscopic Measurement on a Neutron Beam

Setting ω0 = −γB0 and ω1 = −γB1 , write the matrix representation of ˆ H(t) in the basis {|+, |−}. 4.1.2. Treating B 1 as a perturbation, calculate, in ﬁrst order time-dependent perturbation theory, the probability of ﬁnding the neutron in the state |+ at time t = +∞ (far from the interaction zone) if it was in the state |− at t = −∞. One measures the ﬂux of neutrons which have ﬂipped their spins, and are in the state |+ when they leave the ﬁeld zone. This ﬂux is proportional to the probability P−+ that they have undergone the above transition. Show that this probability has a resonant behavior as a function of the applied angular frequency ω. Plot P−+ as a function of the distance from the resonance ω − ω0 . How does the width of the resonance curve vary with v and a? The existence of this width puts a limit on the accuracy of the measurement of ω0 , and therefore of γ. Is there an explanation of this on general grounds? 4.1.3. On the path of the beam, one adds a second zone with an oscillating ﬁeld B 1 . This second zone is identical to the ﬁrst but is translated along the x axis by a distance b (b a):

= B1 e−r /a cos ω(t − z/c) B1x = B1 e−r /a sin ω(t − z/c) B1y =0, B1z

(4.2)

1/2

where r = ((x − b) + y 2 ) . Show that the transition probability P−+ across the two zones can be expressed in a simple way in terms of the transition probability calculated in the previous question. Why is it preferable to use a setup with two zones separated by a distance b rather than a single zone, as in question 4.1.2, if one desires a good accuracy in the measurement of the angular frequency ω0 ? What is the order of magnitude of the improvement in the accuracy? 2

4.1.4. What would be the probability P−+ if one were to use N zones equally spaced by the same distance b from one another? What optical system is this reminiscent of? 4.1.5. Suppose now that the neutrons, still in the initial spin state |−, propagate along the z axis instead of the x axis. Suppose that the length of the interaction zone is b, i.e. that the oscillating ﬁeld is given by (4.1) for −b/2 ≤ z ≤ +b/2 and is zero for |z| > b/2. Calculate the transition probability in this new conﬁguration. P−+ For what value of ω is this probability maximum? Explain the diﬀerence with the result obtained in question 4.1.2.

4.2 Solutions

49

4.1.6. In practice, the neutron beam has some velocity dispersion around the value v. Which of the two methods described in questions 4.1.3 and 4.1.5 is preferable? 4.1.7. Numerical Application: The neutrons of the beam have a de Broglie A. Calculate their velocity. wavelength λn = 31 ˚ In order to measure the neutron gyromagnetic ratio γn , one proceeds as in question 4.1.3. One can assume that the accuracy is given by δω0 =

πv . 2b

The most accurate value of the neutron gyromagnetic ratio is currently γn = −1.912 041 84 (±8.8 × 10−7 ) q/Mp where q is the unit charge and Mp the proton mass. In a ﬁeld B0 = 1 T, what must be the length b in order to achieve this accuracy?

4.2 Solutions 4.1.1. The magnetic Hamiltonian is ˆ ˆ H(t) = −µ.B = −γ B0 Sˆz + B1x (t) Sˆx + B1y (t) Sˆy . Since x = vt, y = z = 0, ˆ H(t) = −γ B0 Sˆz + B1 e−v|t|/a Sˆx cos ωt + Sˆy sin ωt whose matrix representation is

h ¯ ω1 exp(−v|t|/a − iωt) ω0 ˆ H(t) = . −ω0 2 ω1 exp(−v|t|/a + iωt) 4.1.2. Let |ψ(t) = α(t)|+ + β(t)|− be the neutron state at time t. The Schr¨ odinger equation gives the evolution of α and β: ω1 −iωt−v|t|/a ω0 α+ e β 2 2 ω0 ω1 iωt−v|t|/a e β. α− iβ˙ = 2 2

iα˙ =

˜ We now introduce the variables α ˜ and β: α ˜ (t) = α(t) eiω0 t/2 whose evolution is given by

˜ = β(t) e−iω0 t/2 , β(t)

50

4 Spectroscopic Measurement on a Neutron Beam

ω1 i(ω0 −ω)t−v|t|/a ˜ e β 2 ω1 i(ω−ω0 )t−v|t|/a ˙ iβ˜ = e α ˜. 2

iα ˜˙ =

The equation for α ˜ can be formally integrated and it gives ω1 t i(ω0 −ω)t −v|t |/a ˜ α ˜ (t) = β(t ) dt , e 2i −∞

(4.3)

where we have used the initial condition α ˜ (−∞) = α(−∞) = 0. Now, since ˜ ) by its we want the value of α(t) to ﬁrst order in B1 , we can replace β(t ˜ unperturbed value β(t ) = 1 in the integral. This gives ω1 +∞ i(ω0 −ω)t −v|t |/a e dt 2i −∞ 1 ω1 v = . ia (ω − ω0 )2 + (v/a)2

γ−+ ≡ α ˜ (+∞) =

The transition probability is therefore P−+ =

1 ω12 v 2 . a2 [(ω0 − ω)2 + (v/a)2 ]2

Fig. 4.1. Transition probability in one zone

The width of the resonance curve is of the order of v/a. This quantity is the inverse of the time τ = a/v a neutron spends in the oscillating ﬁeld. From the uncertainty relation δE.τ ∼ ¯ h, when an interaction lasts a ﬁnite time τ the accuracy of the energy measurement δE is bounded by δE ≥ ¯h/τ . Therefore, from ﬁrst principles, one expects that the resonance curve will have a width of the order of h ¯ /τ in energy, or 1/τ in angular frequency.

4.2 Solutions

51

4.1.3. In the two-zone case, the transition amplitude (in ﬁrst order perturbation theory) becomes

+∞ +∞ ω1 ei(ω0 −ω)t−v|t|/a dt + ei(ω0 −ω)t−|vt−b|/a dt . γ−+ = 2i −∞ −∞ If we make the change of variables t = t − b/v in the second integral, we obtain +∞ ω1 ei(ω0 −ω)t−v|t|/a dt , 1 + ei(ω0 −ω)b/v γ−+ = 2i −∞ which is the same formula as previously but multiplied by 1 + ei(ω0 −ω)b/v . If we square this expression, in order to ﬁnd the probability, we obtain

1 4ω12 v 2 (ω0 − ω)b 2 cos P−+ = . a2 [(ω0 − ω)2 + v 2 /a2 ]2 2v

Fig. 4.2. Ramsey fringes in a two-zone setup

The envelope of this curve is, up to a factor of 4, the same as the previous curve. However, owing to the extra oscillating factor, the half-width at halfmaximum of the central peak is now of order πv/(2b). The parameter which now governs the accuracy is the total time b/v that the neutron spends in the apparatus, going from one zone to the other. In spectroscopic measurements, it is important to locate the exact position of the maximum of the peak. Multiplying the width of the peak by a factor a/b ( 1 since a b) results in a major improvement of the measurement accuracy. Of course one could in principle build a single interaction zone of large size ∼ b, but it would be diﬃcult to maintain a well controlled oscillating ﬁeld over such a large region. From a practical point of view, it is much simpler to use small interaction zones of size a and to separate them by a large distance b.

52

4 Spectroscopic Measurement on a Neutron Beam

4.1.4. It is quite straightforward to generalize the previous results to an arbitrary number of zones: ω1 γ−+ = 1 + ei(ω0 −ω)b/v + · · · + ei(N −1)(ω0 −ω)b/v 2i +∞ ei(ω0 −ω)t−v|t|/a dt × −∞

P−+

1 sin2 [N (ω0 − ω)b/2v] ω2 v2 = 12 . 2 2 a [(ω0 − ω)2 + v 2 /a2 ] sin [(ω0 − ω)b/2v]

As far as amplitudes are concerned, there is a complete analogy with a diffraction grating in optics. The neutron (more generally, the particle or the atom) has some transition amplitude t for undergoing a spin ﬂip in a given interaction zone. The total amplitude T is the sum T = t + t eiφ + t e2iφ + . . . , where eiφ is the phase shift between two zones. 4.1.5. We now set z = vt, and x = y = 0 for the neutron trajectory. This will modify the phase of the ﬁeld (Doppler eﬀect) ω(t − z/c) → ω(1 − v/c)t = ω ˜t

with

ω ˜ = ω (1 − v/c)

and we must integrate the evolution of α: ˜ iα ˜˙ =

ω1 i(ω0 −˜ω)t ˜ e β 2

(with β˜ 1) between ti = −b/(2v) and tf = b/(2v). The transition probability is then sin2 [(ω0 − ω ˜ )b/(2v)] = ω12 , P−+ 2 (ω0 − ω ˜) which has a width of the order of b/v but is centered at ω ˜ = ω0 ⇒ ω =

ω0 ω0 (1 + v/c) . 1 − v/c

Comparing with question 4.1.2, we ﬁnd that the resonance frequency is displaced: The neutron moves in the propagation direction of the ﬁeld, and there is a ﬁrst order Doppler shift of the resonance frequency. 4.1.6. If the neutron beam has some velocity dispersion, the experimental result will be the same as calculated above, but smeared over the velocity distribution. In the method of question 4.1.3, the position of lateral fringes, and the width of the central peak, vary with v. A velocity distribution will lead to

4.2 Solutions

53

a broader central peak and lateral fringes of decreasing amplitude. However the position of the central peak does not depend on the velocity, and it is therefore not shifted if the neutron beam has some velocity dispersion. On the contrary, in the method of question 4.1.5, the position of the central peak depends directly on the velocity. A dispersion in v will lead to a corresponding dispersion of the position of the peak we want to measure. The ﬁrst method is highly preferable. A, v = h/(Mn λn ) 128 m/s. 4.1.7. Numerically, for λn = 31 ˚ Experimentally, one obtains an accuracy δω0 /ω0 = δγn /γn = 4.6 × 10−7 . For B = 1 T, the angular frequency is ω0 = γn B0 1.8 × 108 s−1 , which gives δω0 /(2π) 13 Hz and b 2.4 m. Actually, one can improve the accuracy considerably by analysing the shape of the peak. In the experiment reported in the reference quoted below, the length b is 2 m and the ﬁeld is B0 = 0.05 T (i.e. an angular frequency 20 times smaller than above). Reference G.L. Greene, N.F. Ramsey, W. Mampe, J.M. Pendlebury, K.F. Smith, W.D. Dress, P.D. Miller, and P. Perrin, Phys. Rev. D 20, 2139 (1979).

5 Analysis of a Stern–Gerlach Experiment

We analyze a Stern–Gerlach experiment, both experimentally and from the theoretical point of view. In the experimental setup considered here, a monochromatic beam of neutrons crosses a region of strongly inhomogeneous magnetic ﬁeld, and one observes the outgoing beam.

5.1 Preparation of the Neutron Beam Neutrons produced in a reactor are ﬁrst “cooled”, i.e. slowed down by crossing liquid hydrogen at 20 K. They are incident on a monocrystal, for instance graphite, from which they are diﬀracted. To each outgoing direction, there corresponds a well-deﬁned wavelength, and therefore a well-deﬁned momentum. A beryllium crystal acts as a ﬁlter to eliminate harmonics, and the vertical extension of the beam is controlled by two gadolinium blocks, which are opaque to neutrons, separated by a thin sheet of (transparent) aluminum of thickness a, which constitutes the collimating slit, as shown in Fig. 5.1.

Fig. 5.1. Preparation of the neutron beam

5.1.1. The de Broglie wavelength of these monochromatic neutrons is λ = 4.32 ˚ A. What are their velocity and their kinetic energy?

56

5 Analysis of a Stern–Gerlach Experiment

5.1.2. One observes the impacts of the neutrons on a detector at a distance L = 1 m from the slit. The vertical extension of the beam at the detector is determined by two factors, ﬁrst the width a of the slit, and second the diﬀraction of the neutron beam by the slit. We recall that the angular width θ of the diﬀraction peak from a slit of width a is related to the wavelength λ by sin θ = λ/a. For simplicity, we assume that the neutron beam is well collimated before the slit, and that the vertical extension δ of the beam on the detector is the sum of the width a of the slit and the width of the diﬀraction peak. Show that one can choose a in an optimal way in order to make δ as small as possible. What is the corresponding width of the beam on the detector? 5.1.3. In the actual experiment, the chosen value is a = 5 µm. What is the observed width of the beam at the detector? Comment on the respective eﬀects of the slit width a and of diﬀraction, on the vertical shape of the observed beam on the detector? The extension of the beam corresponds to the distribution of neutron impacts along the z axis. Since the purpose of the experiment is not only to observe the beam, but also to measure its “position” as deﬁned by the maximum of the distribution, what justiﬁcation can you ﬁnd for choosing a = 5 µm? Figure 5.2 is an example of the neutron counting rate as a function of z. The horizontal error bars, or bins, come from the resolution of the measuring apparatus, the vertical error bars from the statistical ﬂuctuations of the number of neutrons in each bin. The curve is a best ﬁt to the experimental points. Its maximum is determined with an accuracy δz ∼ 5 µm.

Fig. 5.2. Measurement of the beam proﬁle on the detector

5.3 The Stern–Gerlach Experiment

57

5.2 Spin State of the Neutrons In order to completely describe the state of a neutron, i.e. both its spin state and its spatial state, we consider the eigenbasis of the spin projection along the z axis, Sˆz , and we represent the neutron state as

ψ+ (r, t) , |ψ(t) : ψ− (r, t) where the respective probabilities of ﬁnding the neutron in the vicinity of h/2 are point r with its spin component Sz = ±¯ d3 P (r, Sz = ±¯ h/2, t) = |ψ± (r, t)|2 d3 r . 5.2.1. What are the probabilities P± (t) of ﬁnding, at time t, the values ±¯ h/2 when measuring Sz irrespective of the position r? 5.2.2. What is, in terms of ψ+ and ψ− , the expectation value of the x component of the neutron spin Sx in the state |ψ(t)? 5.2.3. What are the expectation values of the neutron’s position r and momentum p in the state |ψ(t)? 5.2.4. We assume that the state of the neutron can be written:

α+ |ψ(t) : ψ(r, t) , α− where the two complex numbers α± are such that |α+ |2 + |α− |2 = 1. How do the results of questions 5.2.2 and 5.2.3 simplify in that case?

5.3 The Stern–Gerlach Experiment Between the slit, whose center is located at the origin (x = y = z = 0), and the detector, located in the plane x = L, we place a magnet of length L whose ﬁeld B is directed along the z axis. The magnetic ﬁeld varies strongly with z; see Fig. 5.3. We assume that the components of the magnetic ﬁeld are Bx = By = 0

Bz = B0 + b z .

In what follows we choose1 B0 = 1 T and b = 100 T/m. 1

This form violates Maxwell’s equation ∇ · B = 0, but it simpliﬁes the following calculation. With a little modiﬁcation (e.g. Bx = 0, By = −b y and By Bz over the region of space crossed by the neutron beam), one can settle this matter, and arrive at the same conclusions.

58

5 Analysis of a Stern–Gerlach Experiment

Fig. 5.3. Magnetic ﬁeld setup in the Stern–Gerlach experiment

ˆ in the matrix representation that The magnetic moment of the neutron µ we have chosen for |ψ is ˆ = µ0 σ ˆ , µ ˆ are the usual Pauli matrices, and µ0 = 1.913 µN , where µN is the where σ h/2Mp = 5.051 10−27 J · T−1 . Hereafter, we denote nuclear magneton µN = q¯ the neutron mass by m. 5.3.1. What is the form of the Hamiltonian for a neutron moving in this magnetic ﬁeld? Write the time-dependent Schr¨ odinger equation for the state |ψ(t). Show that the Schr¨ odinger equation decouples into two equations of the Schr¨ odinger type, for ψ+ and ψ− respectively. 5.3.2. Show that one has d dt

|ψ± (r, t)|2 d3 r = 0 .

What does one conclude as to the probabilities of measuring µz = ±µ0 ? 5.3.3. We assume that, at t = 0, at the entrance of the ﬁeld zone, one has

α+ |ψ(0) : ψ(r, 0) α− and that r = 0, py = pz = 0 and px = p0 = h/λ, where the value of the wavelength λ has been given above. The above conditions correspond to the experimental preparation of the neutron beam discussed in Sect. 5.1. Let Aˆ be an observable depending on the position operator rˆ and the ˆ . We deﬁne the numbers A+ and A− by momentum operator p 1 ∗ ∗ A± = (r, t) Aˆ ψ± (r, t) d3 r . ψ± |α± |2

5.4 Solutions

59

What is the physical interpretation of A+ and A− ? Show in particular that |ψ+ |2 /|α+ |2 and |ψ− |2 /|α− |2 are probability laws. 5.3.4. Apply Ehrenfest’s theorem to calculate the following quantities: d r ± , dt

d p± . dt

Solve the resulting equations and give the time evolution of r ± and p± . Give the physical interpretation of the result, and explain why one observes a splitting of the initial beam into two beams of relative intensities |α+ |2 and |α− |2 . 5.3.5. Calculate the splitting between the two beams when they leave the magnet. Express the result in terms of the kinetic energy of the incident neutrons (we recall that L = 1 m and b = 100 T/m). Given the experimental error δz in the measurement of the position of the maximum intensity of a beam, i.e. δz = 5 × 10−6 m as discussed in question 5.1.3, what is the accuracy on the measurement of the neutron magnetic moment in such an experiment, assuming that the determination of the magnetic ﬁeld and the neutron energy is not a limitation? Compare with the result of magnetic resonance experiments: µ0 = (−1.91304184 ± 8.8 × 10−7 ) µN . 5.3.6. In the same experimental setup, what would be the splitting of a beam of silver atoms (in the original experiment of Stern and Gerlach, the atomic beam came from an oven at 1000 K) of energy E 1.38 × 10−20 J? The magnetic moment of a silver atom is the same as that of the valence electron h/2me 9.3 × 10−24 J.T−1 . |µe | = q¯ 5.3.7. Show that, quite generally, in order to be able to separate the two outgoing beams, the condition to be satisﬁed is of the form E⊥ t ≥ ¯ h/2 , where E⊥ is the transverse kinetic energy acquired by the neutrons in the process, and t is the time they spend in the magnetic ﬁeld. Comment and conclude.

5.4 Solutions Section 5.1: Preparation of the Neutron Beam 5.1.1. We have v = h/(λm) and E = mv 2 /2, which yields v = 916 m s−1 and E = 0.438 × 10−2 eV.

60

5 Analysis of a Stern–Gerlach Experiment

5.1.2. The contribution of diﬀraction to the beam width is δdiﬀ = L tan θ ∼ Lλ/a. With the simple additive prescription (which can be improved, but this would not yield √ very diﬀerent results), we obtain δ = a + Lλ/a which is minimal for a = Lλ 21 µm. The spreading√of the beam on the detector is then equal to the Heisenberg minimum δ = 2 Lλ = 42 µm. The uncertainty relations forbid δ to be less than some lower limit. In other words, the spreading of the wave packet, which increases as a decreases competes with the spatial deﬁnition of the incoming beam. 5.1.3. For a = 5 µm, we have δ = 91.5 µm. In that case, the eﬀect of diﬀraction is predominant. The reason for making this choice is that the shape of the diﬀraction peak is known and can be ﬁtted quite nicely. Therefore, this is an advantage in determining the position of the maximum. However, one cannot choose a to be too small, otherwise the neutron ﬂux becomes too small, and the number of events is insuﬃcient. Section 5.2: Spin State of the Neutrons 5.2.1. P± (t) = |ψ± (r, t)|2 d3 r N.B. The normalization condition (total probability equal to 1) is

P+ + P − = 1 ⇒ |ψ+ (r, t)|2 + |ψ− (r, t)|2 d3 r = 1 . The quantity |ψ+ (r, t)|2 + |ψ− (r, t)|2 is the probability density of ﬁnding the neutron at point r. h/2)ψ|ˆ σx |ψ 5.2.2. By deﬁnition, the expectation value of Sx is Sx = (¯ therefore ∗

h ¯ ∗ ψ+ (r, t)ψ− (r, t) + ψ− Sx = (r, t)ψ+ (r, t) d3 r . 2 5.2.3. Similarly

r = r |ψ+ (r, t)|2 + |ψ− (r, t)|2 d3 r ,

∗ h ¯ ∗ p = (r, t) ∇ψ− (r, t) d3 r . ψ+ (r, t) ∇ψ+ (r, t) + ψ− i 5.2.4. If the variables are factorized, we have the simple results:

∗ Sx = h ¯ Re α+ α−

and r =

r |ψ(r, t)|2 d3 r ,

p =

¯ h i

ψ ∗ (r, t) ∇ψ(r, t) d3 r .

5.4 Solutions

61

Section 5.3: The Stern–Gerlach Experiment 5.3.1. The matrix form of the Hamiltonian is

ˆ2 1 0 p 1 0 ˆ H= . − µ0 (B0 + b zˆ) 0 −1 2m 0 1 The Schr¨ odinger equation is i¯ h

d ˆ |ψ(t) = H|ψ(t) . dt

If we write it in terms of the coordinates ψ± we obtain the uncoupled set h2 ¯ ∂ ψ+ (r, t) = − ∆ψ+ − µ0 (B0 + b z) ψ+ ∂t 2m ∂ h2 ¯ ∆ψ− + µ0 (B0 + b z) ψ− i¯ h ψ− (r, t) = − ∂t 2m i¯ h

d ˆ ± |ψ± , with or, equivalently i¯ h dt |ψ± = H 2

h ˆ± = − ¯ H ∆ ∓ µ0 (B0 − b z) . 2m In other words, we are dealing with two uncoupled Schr¨ odinger equations, where the potentials have opposite values. This is basically what causes the Stern–Gerlach splitting. ˆ± 5.3.2. Since both ψ+ and ψ− satisfy Schr¨ odinger equations, and since H are both hermitian, we have the usual properties of Hamiltonian evolution for ψ+ and ψ− separately, in particular the conservation of the norm. The probability of ﬁnding µz = ±µ0 , and the expectation value of µz are both time independent. 5.3.3. By deﬁnition, we have |ψ± (r, t)|2 d3 r = |α± |2 , where |α± |2 is time independent. The quantities |ψ+ (r, t)|2 /|α+ |2 and |ψ− (r, t)|2 /|α− |2 are the probability densities for ﬁnding a neutron at poh/2 and Sz = −¯ h/2. sition r with, respectively, Sz = +¯ The quantities A+ and A− are the expectation values of the physical h/2 and Sz = −¯ h/2. quantity A, for neutrons which have, respectively, Sz = +¯ 5.3.4. Applying Ehrenfest’s theorem, one has for any observable d 1 ∗ ˆ H ˆ ± ] ψ± (r, t) d3 r . A± = (r, t) [A, ψ± dt i¯ h|α± |2

62

5 Analysis of a Stern–Gerlach Experiment

Therefore

d r ± = p± /m dt

and

d d d px± = py± = 0 pz± = ±µ0 b . dt dt dt The solution of these equations is straightforward: px± = p0 ,

py± = 0 ,

pz± = ±µ0 b t

p0 t µ0 b t2 = vt , y± = 0 , z± = ± . m 2m Consequently, the expectation values of the vertical positions of the neutrons which have µz = +µ0 and µz = −µ0 diverge as time progresses: there is a separation in space of the support of the two wave functions ψ+ and ψ− . The intensities of the two outgoing beams are proportional to |α+ |2 and |α− |2 . x± =

5.3.5. As the neutrons leave the magnet, one has x = L, therefore t = L/v and ∆z = |µ0 b |L2 /mv 2 = |µ0 b |L2 /2E where E is the energy of the incident neutrons. This provides a splitting of ∆z = 0.69 mm. The error on the position of each beam is δz = 5 µm, that is to say a relative error on the splitting of the beams, or equivalently, on the measurement of µ0 √ δµ0 2 δz ∼ 1.5% , µ0 ∆z which is far from the accuracy of magnetic resonance measurements. 5.3.6. For silver atoms, one has |µ0 |/2E = 3.4 × 10−4 T−1 . Hence, in the same conﬁguration, one would obtain, for the same value of the ﬁeld gradient and the same length L = 1 m, a separation ∆z = 3.4 cm, much larger than for neutrons. Actually, Stern and Gerlach, in their ﬁrst experiment, had a much weaker ﬁeld gradient and their magnet was 20 cm long. 5.3.7. The condition to be satisﬁed in order to resolve the two outgoing beams is that the distance ∆z between the peaks should be larger than the full width of each peak (this is a common criterion in optics; by an appropriate inspection of the line shape, one may lower this limit). We have seen in Sect. 5.1 that √ the absolute minimum for the total beam extension√on the detector is 2 Lλ, which amounts to a full width at half-maximum Lλ. In other words, we must have: ∆z 2 ≥ L λ . In the previous section, we have obtained the value of ∆z, and, by squaring, we 2 obtain ∆z 2 = (µ0 b ) t4 /m2 , where t is the time spent traversing the magnet. On the other hand, the transverse kinetic energy of an outgoing neutron is 2 E⊥ = p2z± /(2m) = (µ0 b ) t2 /(2m).

5.4 Solutions

63

Putting the two previous relations together, we obtain ∆z 2 = 2E⊥ t2 /m; inserting this result in the ﬁrst inequality, we obtain E⊥ t ≥ h/2 , where we have used L = vt and λ = h/mv. This is nothing but one of the many forms of the time–energy uncertainty relation. The right-hand side is not the standard h ¯ /2 because we have considered a rectangular shape of the incident beam (and not a Gaussian). This brings in an extra factor of 2π. Physically, this result is interesting in many respects. (a) First, it shows that the eﬀort that counts in making the experiment feasible is not to improve individually the magnitude of the ﬁeld gradient, or the length of the apparatus, etc., but the particular combination of the product of the energy transferred to the system and the interaction time of the system with the measuring apparatus. (b) Secondly, this is a particular example of the fundamental fact stressed by many authors2 that a measurement is never point-like. It has always a ﬁnite extension both in space and in time. The Stern–Gerlach experiment is actually a very good example of a measuring apparatus in quantum mechanics since it transfers quantum information – here the spin state of the neutron – into space–time accessible quantities – here the splitting of the outgoing beams. (c) This time–energy uncertainty relation is encountered in most, if not all quantum measurements. Here it emerges as a consequence of the spreading of the wave packet. It is a simple and fruitful exercise to demonstrate rigorously the above property by calculating directly the time evolution of the following expectation values: 2 pz± 2 − z± 2 , ET = , z± p± + p± z± . z± , ∆z 2 = z± 2m

2

See, for instance, L.D. Landau and E.M. Lifshitz, Quantum Mechanics, Pergamon Press, Oxford, 1965.

6 Measuring the Electron Magnetic Moment Anomaly

In the framework of the Dirac equation, the gyromagnetic factor g of the electron is equal to 2. In other words, the ratio between the magnetic moment and the spin of the electron is gq/(2m) = q/m, where q and m are the charge and the mass of the particle. When one takes into account the interaction of the electron with the quantized electromagnetic ﬁeld, one predicts a value of g slightly diﬀerent from 2. The purpose of this chapter is to study the measurement of the quantity g − 2.

6.1 Spin and Momentum Precession of an Electron in a Magnetic Field Consider an electron, of mass m and charge q (q < 0), placed in a uniform and static magnetic ﬁeld B directed along the z axis. The Hamiltonian of the electron is ˆ 2−µ ˆ = 1 (ˆ ˆ · B, p − q A) H 2m ˆ is the vector potential A ˆ = B × rˆ /2 and µ ˆ is the intrinsic magnetic where A moment operator of the electron. This magnetic moment is related to the ˆ by µ ˆ with γ = (1 + a)q/m. The quantity a is called ˆ = γ S, spin operator S the magnetic moment “anomaly”. In the framework of the Dirac equation, a = 0. Using quantum electrodynamics, one predicts at ﬁrst order in the ﬁne structure constant a = α/(2π). ˆ ˆ = (ˆ The velocity operator is v p − q A)/m, and we set ω = qB/m. 6.1.1. Verify the following commutation relations: ˆ = i¯ hω vˆy ; [ˆ vx , H]

ˆ = −i¯ [ˆ vy , H] hω vˆx ;

ˆ =0. [ˆ vz , H]

6.1.2. Consider the three quantities C1 (t) = Sˆz vˆz ,

C2 (t) = Sˆx vˆx + Sˆy vˆy ,

C3 (t) = Sˆx vˆy − Sˆy vˆx .

66

6 Measuring the Electron Magnetic Moment Anomaly

Write the time evolution equations for C1 , C2 , C3 . Show that these three equations form a linear diﬀerential system with constant coeﬃcients. One will make use of the quantity Ω = a ω. ˆ ·v ˆ ? 6.1.3. What is the general form for the evolution of S 6.1.4. A beam of electrons of velocity v is prepared at time t = 0 in a spin state such that one knows the values of C1 (0), C2 (0), and C3 (0). The beam interacts with the magnetic ﬁeld B during the time interval [0, T ]. One neglects the interactions between the electrons of the beam. At time T , one ˆ v . measures a quantity which is proportional to S.ˆ The result of such a measurement is presented in Fig. 6.1 as a function of the time T , for a value of the magnetic ﬁeld B = 9.4 × 10−3 T (data taken from D.T. Wilkinson and H.R. Crane, Phys. Rev. 130, 852 (1963)). Deduce from this curve an approximate value for the anomaly a. 6.1.5. Does the experimental value agree with the prediction of quantum electrodynamics?

ˆ v , as a function of the time T Fig. 6.1. Variations of the quantity S.ˆ

6.2 Solutions ˆ = mˆ 6.1.1. The electron Hamiltonian is H v 2 /2 − γB Sˆz . The following commutation relations can be established with no diﬃculty hqB/m2 = i¯ hω/m, [ˆ vx , vˆz ] = [ˆ vy , vˆz ] = 0 , [ˆ vx , vˆy ] = i¯ 2 [ˆ vx , vˆy ] = [ˆ vx , vˆy ]ˆ vy + vˆy [ˆ vx , vˆy ] = 2i¯ hω vˆy /m . Therefore ˆ = i¯ [ˆ vx , H] hωˆ vy ;

ˆ = −i¯ [ˆ vy , H] hωˆ vx ;

ˆ =0. [ˆ vz , H]

6.2 Solutions

67

ˆ = [O, ˆ H], ˆ 6.1.2. We make use of the property i¯h(d/dt)O valid for any observable (Ehrenfest theorem). The time evolution of C1 is trivial: ˆ =0 ⇒ [Sˆz vˆz , H]

dC1 =0; dt

C1 (t) = A1 ,

where A1 is a constant. For C2 and C3 , we proceed in the following way: ˆ = [Sˆx vˆx , mˆ [Sˆx vˆx , H] v 2 /2] − γB[Sˆx vˆx , Sˆz ] = i¯ hω(Sˆx vˆy + (1 + a)Sˆy vˆx ). Similarly, ˆ = −i¯ [Sˆy vˆy , H] hω(Sˆy vˆx + (1 + a)Sˆx vˆy ) ˆ = −i¯ [Sˆx vˆy , H] hω(Sˆx vˆx − (1 + a)Sˆy vˆy ) ˆ = i¯ [Sˆy vˆx , H] hω(Sˆy vˆy − (1 + a)Sˆx vˆx ) . Therefore, ˆ = −i¯ [Sˆx vˆx + Sˆy vˆy , H] hωa (Sˆx vˆy − Sˆy vˆx ) ˆ = i¯ [Sˆx vˆy − Sˆy vˆx , H] hωa(Sˆx vˆx + Sˆy vˆy ) and

dC2 = −ΩC3 , dt

dC3 = ΩC2 . dt

6.1.3. We therefore obtain d2 C2 /dt2 = −Ω 2 C2 , whose solution is C2 (t) = A2 cos (Ωt + ϕ) , where A2 and ϕ are constant. Hence, the general form of the evolution of S · v is S · v(t) = C1 (t) + C2 (t) = A1 + A2 cos (Ωt + ϕ) . In other words, in the absence of anomaly, the spin and the momentum of the electron would precess with the same angular velocity: the cyclotron frequency (precession of momentum) and the Larmor frequency (precession of magnetic moment) would be equal. Measuring the diﬀerence in these two frequencies gives a direct measurement of the anomaly a, of fundamental importance in quantum electrodynamics. 6.1.4. One calculates the anomaly from the relation a = Ω/ω. The experimental results for S · v show a periodic behavior in time with a period τ ∼ 3 µs, i.e. Ω = 2π/τ ∼ 2 × 106 s−1 . In a ﬁeld B = 0.0094 T, ω = 1.65 × 109 s−1 , and a = Ω/ω ∼ 1.2 × 10−3 . 6.1.5. This value is in good agreement with the theoretical prediction a = α/2π = 1.16 × 10−3 .

68

6 Measuring the Electron Magnetic Moment Anomaly

Remark: The value of the anomaly is now known with an impressive accuracy: atheo. = 0.001 159 652 200 (40) aexp. = 0.001 159 652 193 (10) . The theoretical calculation includes all corrections up to order 3 in α.

7 Decay of a Tritium Atom

The nucleus of the tritium atom is the isotope 3 H, of charge Z = 1. This nucleus is radioactive and transforms into a 3 He nucleus by β decay. The purpose of this chapter is to study the electronic state of the 3 He+ ion formed after the decay. We consider nuclei as inﬁnitely massive compared to the electron, of ¯ 2 /(me2 ) for the Bohr radius and EI = mc2 α2 /2 mass m. We write a1 = h 13.6 eV for the ionization energy of the hydrogen atom, where α is the ﬁne structure constant [e2 = q 2 /(4π0 ), where q is the electron charge]. In the ground state |ψ0 of the tritium atom, the wave function of the electron (n = 1, l = 0, m = 0) is the same as in the normal hydrogen atom: 1 ψ0 (r) = √ e−r/a1 . (7.1) πa1 3 The β decay of the tritium nucleus leads to: 3

H → 3 He + e− + ν¯

(7.2)

(¯ ν is an antineutrino), where the emitted electron has an energy of the order of 15 keV and the helium nucleus 3 He has charge Z = 2. The decay is an instantaneous process; the β electron is emitted with a large velocity and leaves the atomic system very rapidly. Consequently, an ionized 3 He+ atom is formed, for which, at the time t0 of the decay, the wave function of the electron is practically the same as in tritium, and we shall assume it is still given by (7.1). We denote by |n, l, m the states of the ionized helium atom which is a hydrogen-like system, i.e. one electron placed in the Coulomb ﬁeld of a nucleus of charge 2.

7.1 The Energy Balance in Tritium Decay ˆ 1 of the atomic electron before the decay and 7.1.1. Write the Hamiltonian H ˆ the Hamiltonian H2 of this electron after the decay (when the potential term has suddenly changed).

70

7 Decay of a Tritium Atom

7.1.2. What are, in terms of EI , the energy levels of the 3 He+ atom? Give its Bohr radius and its ground state wave function ϕ100 (r). 7.1.3. Calculate the expectation value E of the energy of the electron after the decay. One can for instance make use of the fact that: ψ0 |

1 1 |ψ0 = r a1

2 ˆ2 = H ˆ1 − e . H r

and

Give the value of E in eV. 7.1.4. Express in terms of |ψ0 and |n, l, m the probability amplitude c(n, l, m) and the probability p(n, l, m) of ﬁnding the electron in the state |n, l, m of 3 He+ after the decay. Show that only the probabilities pn = p (n, 0, 0) do not vanish. 7.1.5. Calculate the probability p1 of ﬁnding the electron in the ground state of 3 He+ . What is the corresponding contribution to E? 7.1.6. A numerical calculation gives the following values: p2 =

1 , 4

∞

pn = 0.02137,

n=3

∞ pn = 0.00177. 2 n n=3

∞ Calculate the probability n=1 pn of ﬁnding the atomic electron in a bound state of 3 He+ and the corresponding contribution to E. Comment on the result. 7.1.7. Experimentally, in the β decay of the tritium atom, one observes that, in about 3% of the events, there are two outgoing electrons, one with a mean kinetic energy Ek ∼ 15 keV, the other with Ek ∼ 34.3 eV, thus leaving a completely ionized 3 He2+ nucleus, as if the β decay electron had “ejected” the atomic electron. Explain this phenomenon.

7.2 Solutions 7.1.1. The two Hamiltonians are 2 2 ˆ 1 = pˆ − e H 2m r

2 2 ˆ 2 = pˆ − 2e . H 2m r

7.1.2. The energy levels corresponding to the bound states of a hydrogenlike atom of nuclear charge Z are En = −Z 2 EI /n2 . In the present case, En = −4EI /n2 . The new Bohr radius is a2 = a1 /2, and the wave function is 1 ϕ100 (r) = 3 e−r/a2 . πa2

7.3 Comments

71

7.1.3. The expectation value of the electron energy in the new nuclear conﬁguration is 2 ˆ 2 |ψ0 = ψ0 |H ˆ 1 |ψ0 − ψ0 | e |ψ0 , E = ψ0 |H r which amounts to

E = −EI −

e2 = −3EI −40.8 eV . a1

7.1.4. By deﬁnition, the probability amplitude is c(n, l, m) = n, l, m|ψ0 , and the probability p(n, l, m) = |n, l, m|ψ0 |2 . The analytic form is ∗ c(n, l, m) = Rnl (r) (Yl,m (θ, φ)) ψ0 (r) d3 r , where Rnl (r) are the radial wave functions of the 3 He+ hydrogen-like atom. Since ψ0 is of the form ψ0 (r) = χ(r)Y0,0 (θ, φ), the orthogonality of spherical harmonics implies p(n, l, m) = 0 if (l, m) = (0, 0). 7.1.5. The probability amplitude in the lowest energy state is √ −r/a2 −r/a1 e e 16 2 1/2 2 (p1 ) . = 4π r dr = 27 πa32 πa31 Hence the probability p1 = 0.70233 and the contribution to the energy p1 E1 = −38.2 eV. 7.1.6. With the numerical ∞ values given in the text, one has p2 E2 = −EI /4 = −3.4 eV, and p = 1 pn = 0.9737 . The contribution to E is EB = ∞ p E = −3.0664 E n n I = −41.7 eV. 1 The total probability is smaller than 1; there exists a non-zero probability (1 − p) = 0.026 that the atomic electron is not bound in the ﬁnal state. The contribution of bound states EB = −41.7 eV is smaller than the total expectation value of the energy E by 0.9 eV. The probability (1 − p) corresponds therefore to a positive electron energy, i.e. an ionization of 3 He+ into 3 He2+ with emission of the atomic electron. 7.1.7. There is necessarily a probability 1 − p = 0.026 for the atomic electron not to be bound around the helium nucleus, therefore that the helium atom be completely ionized in the decay. If the mean kinetic energy of the expelled electron is Ek ∼ 34.3 eV, this represents a contribution of the order of (1 − p)Ek ∼ +0.89 eV to the mean energy which compensates the apparent energy deﬁcit noted above.

7.3 Comments This type of reaction is currently being studied in order to determine the neutrino mass. If M1 and M2 are the masses of the two nuclei, Eβ the energy

72

7 Decay of a Tritium Atom

of the β electron, E the energy of the atomic electron, and Eν¯ the neutrino energy, energy conservation gives for each event: M1 c2 − EI = M2 c2 + Eβ + Eν¯ + E. For a given value of E, the determination of the maximum energy of the β electron (which covers all the spectrum up to 19 keV in the tritium atom case) provides a method for determining the minimum value mν¯ c2 of Eν¯ through this energy balance. An important theoretical problem is that current experiments are performed on molecular tritium (HT or TT molecules) and that molecular wave functions are not known explicitly, contrary to the atomic case considered here. The most precise experiment up to date is reported in Weinheimer et al., Phys. Lett. B460, 219, (1999).

8 The Spectrum of Positronium

The positron e+ is the antiparticle of the electron. It is a spin-1/2 particle, which has the same mass m as the electron, but an electric charge of opposite sign. In this chapter we consider the system called positronium which is an atom consisting of an e+ e− pair.

8.1 Positronium Orbital States We ﬁrst consider only the spatial properties of the system, neglecting all spin eﬀects. We only retain the Coulomb interaction between the two particles. No proof is required, an appropriate transcription of the hydrogen atom results suﬃces. 8.1.1. Express the reduced mass of the system µ, in terms of the electron mass m. 8.1.2. Write the Hamiltonian of the relative motion of the two particles in terms of their separation r and their relative momentum p. 8.1.3. What are the energy levels of the system, and their degeneracies? How do they compare with those of hydrogen? 8.1.4. What is the Bohr radius a0 of the system? How do the sizes of hydrogen and positronium compare? 8.1.5. Give the expression for the normalized ground state wave function ¯, ψ100 (r). Express |ψ100 (0)|2 in terms of the fundamental constants: m, c, h and the ﬁne structure constant α.

8.2 Hyperﬁne Splitting We now study the hyperﬁne splitting of the ground state.

74

8 The Spectrum of Positronium

8.2.1. What is the degeneracy of the orbital ground state if one takes into account spin variables (in the absence of a spin–spin interaction)? 8.2.2. Explain why the (spin) gyromagnetic ratios of the positron and of the electron have opposite signs: γ1 = −γ2 = γ. Express γ in terms of q and m. 8.2.3. One assumes that, as in hydrogen, the spin–spin Hamiltonian in the orbital ground state is: ˆ2 , ˆ1 · S ˆ SS = A S H (8.1) h2 ¯ where the constant A has the dimension of an energy. ˆ SS in the spin basis {|σ1 , σ2 }, Recall the eigenstates and eigenvalues of H where σ1 = ±1, σ2 = ±1. 8.2.4. As in hydrogen, the constant A originates from a contact term: A=−

2 1 γ1 γ2 h ¯ 2 |ψ100 (0)|2 . 3 0 c2

(8.2)

(a) The observed hyperﬁne line of positronium has a frequency ν 200 GHz, compared to ν 1.4 GHz for hydrogen. Justify this diﬀerence of two orders of magnitude. (b) Express the constant A in terms of the ﬁne structure constant and the energy mc2 . Give the numerical value of A in eV. (c) What frequency of the hyperﬁne transition corresponds to this calculated value of A? 8.2.5. Actually, the possibility that the electron and the positron can annihiˆ A in the hyperﬁne Hamiltonian. One late, leads to an additional contribution H ˆ can show that HA does not aﬀect states of total spin equal to zero (S = 0), and that it increases systematically the energies of S = 1 states by the amount: ˆA : H

δE S=1 =

3A 4

(δE S=0 = 0) ,

(8.3)

where A is the same constant as in (8.2). (a) What are the energies of the S = 1 and S = 0 states, if one takes into account the above annihilation term? (b) Calculate the frequency of the corresponding hyperﬁne transition.

8.3 Zeeman Eﬀect in the Ground State The system is placed in a constant uniform magnetic ﬁeld B directed along the z axis. The additional Zeeman Hamiltonian has the form ˆ Z = ω1 Sˆ1z + ω2 Sˆ2z , H where ω1 = −γ1 B and ω2 = −γ2 B.

8.4 Decay of Positronium

75

8.3.1. (a) Taking into account the result of question 2.2 and setting ω = ˆ Z on the basis states {|σ1 , σ2 }. −γB, write the action of H (b) Write in terms of A and h ¯ ω the matrix representation of ˆA + H ˆZ ˆ =H ˆ SS + H H

(8.4)

in the basis {|S, m} of the total spin of the two particles. (c) Give the numerical value of ¯hω in eV for a ﬁeld B = 1 T. Is it easy experimentally to be in a strong ﬁeld regime, i.e. h ¯ ω A? 8.3.2. Calculate the energy eigenvalues in the presence of the ﬁeld B; express the corresponding eigenstates in the basis {|S, m} of the total spin. The largest eigenvalue will be written E+ and the corresponding eigenstate |ψ+ . For convenience, one can√introduce the quantity hω/(7A), and the angle √ x = 8¯ θ deﬁned by sin 2θ = x/ 1 + x2 , cos 2θ = 1/ 1 + x2 . 8.3.3. Draw qualitatively the variations of the energy levels in terms of B. Are there any remaining degeneracies?

8.4 Decay of Positronium We recall that when a system A is unstable and decays: A → B + · · · , the probability for this system to decay during the interval [t, t+dt] if it is prepared at t = 0, is dp = λe−λt dt, where the decay rate λ is related to the lifetime τ of the system by τ = 1/λ. If the decay can proceed via diﬀerent channels, e.g. A → B + · · · and A → C + · · · , with respective decay rates λ1 and λ2 , the total decay rate is the sum of the partial rates, and the lifetime of A is τ = 1/(λ1 + λ2 ). In all what follows, we place ourselves in the rest frame of the positronium. 8.4.1. In a two-photon decay, or annihilation, of positronium, what are the energies of the two outgoing photons, and what are their relative directions? 8.4.2. One can show that the annihilation rate of positronium into photons in an orbital state |n, l, m is proportional to the probability for the electron and positron to be at the same point, i.e. to |ψnlm (0)|2 . In what orbital states is the annihilation possible? 8.4.3. In quantum ﬁeld theory, one can show that, owing to charge conjugation invariance, (a) a singlet state, S = 0, can only decay into an even number of photons: 2, 4, · · · (b) a triplet state, S = 1, can only decay into an odd number of photons: 3, 5, · · · In the orbital ground state ψ100 , split by spin–spin interactions as calculated in Sect. 2, the lifetime of the singlet state is τ2 ∼ 1.25 × 10−10 s, and the

76

8 The Spectrum of Positronium

lifetime of either of the three triplet states is τ3 ∼ 1.4 × 10−7 s. Quantum ﬁeld theory predicts:

2 h ¯ 1 1 4 2 (π − 9) α λ2 . = 4πα2 c |ψ100 (0)|2 , λ3 = = λ2 = τ2 mc τ3 9π Compare theory and experiment. 8.4.4. In order to determine the hyperﬁne constant A of positronium, it is of interest to study the energy and the lifetime of the level corresponding to the state |ψ+ , deﬁned in question 3.2, as a function of the ﬁeld B. ¿From now on, we assume that the ﬁeld is weak, i.e. |x| = |8¯hω/(7A)| 1, and we shall make the corresponding approximations. (a) What are, as a function of x, the probabilities pS ans pT of ﬁnding the state |ψ+ in the singlet and triplet states respectively? + (b) Use the result to calculate the decay rates λ+ 2 and λ3 of the state |ψ+ into two and three photons respectively, in terms of the parameter x, and of the rates λ2 and λ3 introduced in question 4.4. (c) What is the lifetime τ + (B) of the state |ψ+ ? Explain qualitatively its dependence on the applied ﬁeld B, and calculate τ + (B) for B = 0.4 T . 1. One measures, as a function of B, the ratio R = τ + (B)/τ + (0) of the lifetime of the |ψ+ state with and without a magnetic ﬁeld. The dependence on B of R is given in Fig. 8.1, with the corresponding error bars. (i) What estimate does one obtain for the hyperﬁne constant, A, using the value of the magnetic ﬁeld for which the ratio R has decreased by a factor two? (ii) How do theory and experiment compare?

Fig. 8.1. Variation of the ratio R deﬁned in the text as a function of the applied magnetic ﬁeld B

8.5 Solutions

77

8.5 Solutions Section 8.1: Positronium Orbital States In positronium, we have, by scaling: 8.1.1. A reduced mass µ = m/2. ˆ =p ˆ 2 /2µ − q 2 /4π0 r. 8.1.2. A center of mass Hamiltonian H 8.1.3. The energy levels En = −(1/2)µc2 α2 /n2 = −(1/4)mc2 α2 /n2 . The degeneracy is n2 for each level, as in the hydrogen atom; the bound state energies are half of those of hydrogen. ˚ 8.1.4. The Bohr radius is a0 = h ¯ /(µcα) = 2¯ h/(mcα) = 2aH 0 ∼ 1.06 A. The , and, since the proton is ﬁxed, diameter of positronium is r = 3a0 /2 = 3aH 0 the diameter of the hydrogen is 2rH = 3aH . Therefore the two systems have 0 the same size. 8.1.5. The ground state wave function is ψ100 (r) = e−r/a0 / πa30 , and we 3 have |ψ100 (0)|2 = (mcα/(2¯ h)) /π. Section 8.2: Hyperﬁne Splitting 8.2.1. In the orbital ground state, the degeneracy is 4, corresponding to the number of independent spin states. 8.2.2. Since the masses are equal, but the charges are opposite, we have γ1 = q/m, γ2 = −q/m, γ = q/m. 8.2.3. As usual, we can express the spin–spin operator in terms of the total spin S as S 1 · S 2 = [S 2 − S12 − S22 ]/2. Hence, the orbital ground state is split into: √ (a) the triplet states: | + +, (| + − + | − +)/ 2, | − −, with the energy shift: E T = A/4 , √ (b) the singlet state: (| + − − | − +)/ 2, with the energy shift: E S = −3A/4 . 8.2.4. (a) There is a mass factor of ∼ 1/2000, a factor of ∼ 2.8 for the gyromagnetic ratio of the proton, and a factor of 8 due to the value of the wave function at the origin. Altogether, this results in a factor of ∼ 22/2000 ∼ 1% for the ratio of hyperﬁne splittings H/(e+ e− ). (b) The numerical value of A is A=

1 12π0

q¯ h mc

2

mcα 3 1 = mc2 α4 ∼ 4.84 × 10−4 eV. h ¯ 3

78

8 The Spectrum of Positronium

(c) This corresponds to a transition frequency of ν = A/h 117 GHz. This prediction is not in agreement with the experimental result (∼ 200 GHz). ˆ A , the triplet state energy is A while the 8.2.5. (a) Taking into account H singlet state energy is −3A/4. The splitting is δE = 7A/4 = 8.47 × 10−4 eV. (b) The corresponding frequency is ν = δE/h ∼ 205 GHz, in agreement with experiment. Section 8.3: Zeeman Eﬀect in the Ground State 8.3.1. ˆ Z = ω(Sˆ1z − Sˆ2z ), therefore, we have (a) The Zeeman Hamiltonian is H ˆ Z | + + = H ˆ Z | + − = H ˆ HZ | − + =

ˆ Z | − − = 0 H h ¯ ω| + − −¯ hω| − + .

In terms of total spin states, this results in ˆ Z |1, 1 = H ˆ Z |1, −1 = 0 H ˆ Z |1, 0 = h H ¯ ω|0, 0 ˆ Z |0, 0 = h H ¯ ω|1, 0 . (b) Hence the matrix representation in the coupled basis: ⎛

0 0 0 ⎜0 0 0 ˆZ = ⎜ H ⎝0 0 0 0 0 h ¯ω

⎞ 0 0 ⎟ ⎟ , ¯hω ⎠ 0

where the elements are ordered according to: |1, 1, |1, −1, |1, 0, |0, 0. 8.3.2. Similarly, one has the matrix tonian: ⎛ A 0 ⎜0 A ˆ =⎜ H ⎝0 0 0 0

representation of the full spin Hamil⎞ 0 0 0 0 ⎟ ⎟. A ¯hω ⎠ h ¯ ω −3A/4

In a ﬁeld of 1 T, |¯ hω| = q¯ hB/m = 2µB B 1.16 × 10−4 eV. The strong ﬁeld regime corresponds to |¯ hω| A , i.e. B 4 T, which is diﬃcult to reach.

8.5 Solutions

79

8.3.3. (a) Two eigenstates are obvious: |1, 1 and |1, −1, which correspond to the same degenerate eigenvalue A of the energy. The two others are obtained by diagonalizing a 2 × 2 matrix: |ψ+ = cos θ |1, 0 + sin θ |0, 0 , |ψ− = − sin θ |1, 0 + cos θ |0, 0 , corresponding to the energies A E± = ± 8

7A 8

1/2

2 2

+ (¯ hω)

=

A 1 ± 7 1 + x2 . 8

(b) The triplet states | + + and | − − remain degenerate, as shown in Fig. 8.2.

Fig. 8.2. Variation of the hyperﬁne energy levels with applied magnetic ﬁeld

Section 8.4: Decay of Positronium 8.4.1. In a two-photon decay, the outgoing photons have opposite momenta, their energies are both mc2 = 511 keV. 8.4.2. The wave function vanishes at the origin, except for s-waves (|ψnlm (0)|2 = 0 if l = 0), owing to the centrifugal barrier. Therefore the decay can only occur when the positronium is in an s-state. h) which yields τ2 = 8.4.3. The given formulas correspond to λ2 = mc2 α5 /(2¯ 1.24 × 10−10 s and τ3 = 1.38 × 10−7 s, in agreement with experiment.

80

8 The Spectrum of Positronium

8.4.4. (a) For a given value of the applied ﬁeld, with the positronium prepared in the state |ψ+ , the probabilities of ﬁnding the system in the singlet and triplet states are respectively pS = sin2 θ ∼ x2 /4 and pT = cos2 θ ∼ 1 − x2 /4. (b) The rate for |ψ+ to decay into two photons is the product of the probability of ﬁnding |ψ+ in the singlet state with the singlet state decay rate: S 2 2 λ+ 2 = p λ2 ∼ x λ2 /4 = x /(4τ2 ) .

Similarly, one has T 2 2 λ+ 3 = p λ3 ∼ (1 − x /4)λ3 = (1 − x /4)/τ3 .

(c) The lifetime of the |ψ+ state is τ+ =

λ+ 2

1 τ3 τ3 . + = 2 2 τ 2 2 + λ3 1 − x4 + x4 τ32 1 + 16¯h ω2 ττ32 49A

As the ﬁeld B increases, the state |ψ+ , which is purely triplet for B = 0, acquires a greater and greater singlet component. Therefore its lifetime decreases as B increases. For B = 0.4 T, one has τ+ = 0.23 τ3 = 3.2 × 10−8 s. (d) Experimentally, one has R ∼ 0.5, i.e. x2 τ3 /4τ2 1 for B ∼ 0.22 T. hω/7x and h ¯ ω = 2.3 × 10−5 eV, the Therefore x 6 × 10−2 and, since A = 8¯ −4 result is A ∼ 4.4 × 10 eV, in good agreement with theoretical expectations. Section 8.5: References S. DeBenedetti and H.C. Corben, Positronium, Ann. Rev. Nucl. Sci., 4, 191 (1954). Stephan Berko and Hugh N. Pendleton, Positronium, Ann. Rev. Nucl. Sci., 30, 543 (1980). A.P. Mills and S. Chu, Precision Measurements in Positronium, in Quantum Electrodynamics, ed. by T. Kinoshita (World Scientiﬁc, Singapore 1990) pp. 774-821.

9 The Hydrogen Atom in Crossed Fields

We study the modiﬁcation of the energy spectrum of a hydrogen atom placed in crossed static electric and magnetic ﬁelds in perturbation theory. We thus recover a result ﬁrst derived by Pauli. In his famous 1925 paper on the hydrogen atom, W. Pauli made use of the particular symmetry of the Coulomb problem. In addition to the hydrogen spectrum, he was able to calculate the splitting of the levels in an electric ﬁeld (Stark eﬀect) or in a magnetic ﬁeld (Zeeman eﬀect). Pauli also noticed that he could obtain a simple and compact formula for the level splitting in a superposition of a magnetic ﬁeld B 0 and an electric ﬁeld E 0 both static and uniform, and perpendicular to each other. In this case, he found that a level with principal quantum number n is split into 2n − 1 sublevels (k) En + δEn with ¯ k (ω02 + ωe2 )1/2 , (9.1) δEn(k) = h where k is an integer ranging from −(n−1) to n−1, ω0 and ωe are respectively proportional to B0 and E0 , and ωe can be written as ωe =

3 Ωe f (n) 2

with

Ωe =

4π0 ¯h E0 , M qe

where M and qe are the mass and charge of the electron, and where f (n) depends on n only. It is only in 1983 that Pauli’s result was veriﬁed experimentally. Our purpose, here, is to prove (9.1) in the special case n = 2, to calculate ω0 and ωe in that case, and, by examining the experimental result for n = 34, to guess what was the very simple formula found by Pauli for f (n).

82

9 The Hydrogen Atom in Crossed Fields

9.1 The Hydrogen Atom in Crossed Electric and Magnetic Fields We consider the n = 2 level of the hydrogen atom. We neglect all spin eﬀects. We assume that B 0 is along the z axis and E 0 along the x axis. We use ﬁrst order perturbation theory. 9.1.1. What are the energy levels and the corresponding eigenstates in the presence of B 0 only? Check that (9.1) is valid in this case and give the value of ω0 ? 9.1.2. In the presence of E 0 only, the perturbing Hamiltonian is the electric ˆ 0 = −qe rˆ .E 0 . Write the matrix representing H ˆ E in ˆ E = −D.E dipole term H the n = 2 subspace under consideration. We recall √ ∞ that: (a) 0 r3 R2s (r) R2p (r) dr = 3 3 a1 where R2s and R2p are the radial wave functions for the level n = 2, l = 0 and n = 2, l = 1 respectively, and ¯ 2 /(M e2 ) is the Bohr radius (e2 = qe2 /4π0 ). where a1 = h (b) In spherical coordinates (θ polar angle and φ azimuthal angle), the l = 0 and l = 1 spherical harmonics are 3 1 ±1 0 sin θ e±iφ , Y1 (θ, φ) = ∓ Y0 (θ, φ) = √ , 8π 4π 3 0 Y1 (θ, φ) = cos θ. (9.2) 4π 9.1.3. Calculate the energies of the levels originating from the n = 2 level in the presence of the crossed ﬁelds E 0 and B 0 . Show that one recovers (9.1) with ωe = (3/2)f (2)Ωe , and give the value of f (2).

9.2 Pauli’s Result The ﬁrst experimental veriﬁcation of Pauli’s result was performed in 1983.1 In Fig. 9.1, the points correspond to a sub-level with a given value of k arising from the n = 34 level of an hydrogen-like atom. All points correspond to the same energy of this level, but to diﬀerent values of the static ﬁelds E0 and B0 . Knowing that ωe is a function of the principal quantum number n of the form: ωe = (3/2)f (n)Ωe , and that ω0 and Ωe are the constants introduced above, answer the following questions: 1

Fig. 9.1 was obtained by F. Biraben, D. Delande, J.-C. Gay, and F. Penent, with rubidium atoms prepared in a Rydberg state, i.e. with an electron placed in a strongly excited level (see J.-C. Gay, in Atoms in unusual situations, J.-P. Briand ed., p. 107, Plenum, New York, 1986).

9.3 Solutions

83

Fig. 9.1. Values of the electric and magnetic ﬁelds giving rise to the same sub-level energy of the n = 34 level of a hydrogen-like atom

9.2.1. Does the experimental data agree with (9.1)?

9.2.2. Write the quantity ω02 + ωe2 in the form λ γB02 + f 2 (n)E02 , give the value of the constant γ, and calculate f (34). 9.2.3. Guess Pauli’s result concerning f (n).

9.3 Solutions Section 9.1: The Hydrogen Atom in Crossed Electric and Magnetic Fields 9.1.1. Consider a state |n, l, m. The orbital magnetic moment of the elecˆ with γ0 = qe /(2M ). The magnetic Hamiltonian is ˆ orb = γ0 L, tron is µ ˆ ˆ z B0 . ˆ orb .B = −(qe /2M )L H = −µ At ﬁrst order perturbation theory, the energy levels originating from the n = 2 subspace (angular momentum l = 0 or l = 1) are m¯hω0 with m = −1, 0, +1, and ω0 = −qe B0 /(2M ) (ω0 > 0 for B0 > 0). The corresponding states are |2s and |2p, m = 0 |2p, m = −1 |2p, m = +1

δE = 0 δE = −¯ hω0 δE = +¯ hω0 .

ˆ E = −qe x 9.1.2. The Hamiltonian is H ˆ E0 . We have to calculate the 16 matrix x|2, l, m. The integral to be evaluated is elements 2, l , m |ˆ

84

9 The Hydrogen Atom in Crossed Fields

2, l , m |ˆ x|2, l, m =

×

∞

∗ Ylm (θ, φ) sin θ cos φ Ylm (θ, φ) d2 Ω ∗

r3 (R2,l (r)) R2,l (r) dr .

0

The angular integral vanishes if l = l . We need only consider the terms l = 0, l = 1 (and the hermitian conjugate l = 1, l = 0), i.e. √ 2π 1 √ 3 3 a1 (−Y11 (θ, φ) + Y1−1 (θ, φ)) Y1m (θ, φ) d2 Ω 3 4π where we have incorporated √ the radial integral given in the text. One therefore obtains 3 a1 (δm,−1 − δm,1 )/ 2. The only non-vanishing matrix elements are ˆ 2s|H|2p, m = ±1 and their hermitian conjugates. hE0 /(M qe ) = qe E0 a1 /¯ h, we obtain the matrix Setting Ωe = 4π0 ¯ ⎛ ⎞ 0 0 0 1 hΩ ⎜ 0 0 0 0⎟ ˆ E = 3¯ ⎟ . √ e⎜ H 0 −1 ⎠ 2 ⎝0 0 1 0 −1 0 where the rows (columns) are ordered as 2p, m = 1, 0, −1; 2s. 9.1.3. We want to ﬁnd the eigenvalues of the matrix √ ⎞ ⎛ ω0 0 0 3Ωe / 2 ⎜ 0 0 0 0 √ ⎟ ⎟ . h⎜ ¯ ⎝ 0√ 0 −ω0√ −3Ωe / 2 ⎠ 3Ωe / 2 0 −3Ωe / 2 0 There is an obvious eigenvalue λ = 0 since the |2p, m = 0 and |2s states do not mix in the presence the electric ﬁeld. The three other eigenvalues are easily obtained as the solutions of: λ(¯ h2 ω02 − λ2 ) + 9 h ¯ 2 Ωe2 λ = 0 , i.e. λ = 0 and λ = ±¯ h ω02 + 9Ωe2 . The shifts of the energy levels are therefore: δE = 0 twice degenerate, and δE = ±¯ h ω02 + 9Ωe2 . If we adopt the prescription given in the text, we obtain ωe = 3Ωe =⇒ f (2) = 2 . Section 9.2: Pauli’s Result 9.2.1. We remark that the experimental points are aligned on a straight line aB02 + bE02 = constant which is in agreement with (9.1), i.e. a constant value of ω02 + ωe2 corresponds to a constant value of each energy level.

9.3 Solutions

85

9.2.2. Given the deﬁnitions of ω0 and Ωe , one has ω02 + ωe2 =

9 4

h 4π0 ¯ M qe

2

αc 2 2 B0 + f 2 (n) E02 3

,

where α is the ﬁne structure constant and c the velocity of light. The experimental line 2 (αc/3) B02 + f 2 (34)E02 goes through the points E02 = 0, B02 87 × 10−4 T2 and B02 = 0, E02 4 × 106 V2 m−2 . This gives f (34) = 34. 9.2.3. Indeed, the very simple result found by Pauli was f (n) = n.

10 Energy Loss of Ions in Matter

When a charged particle travels through condensed matter, it loses its kinetic energy gradually by transferring it to the electrons of the medium. In this chapter we evaluate the energy loss of the particle as a function of its mass and its charge, by studying the modiﬁcations that the state of an atom undergoes when a charged particle passes in its vicinity. We show how this process can be used to identify the products of a nuclear reaction. The electric potential created by the moving particle appears as a timedependent perturbation in the atom’s Hamiltonian. In order to simplify the problem, we shall consider the case of an atom with a single external electron. The nucleus and the internal electrons will be treated globally as a core of charge +q, inﬁnitely massive and, therefore, ﬁxed in space. We also assume that the incident particle of charge Z1 q is heavy and non-relativistic, and that its kinetic energy is large enough so that in good approximation its motion can be considered linear and uniform, of constant velocity v, when it interacts with an atom. Here q denotes the unit charge and we set e2 = q 2 /(4πε0 ). We consider the x, y plane deﬁned by the trajectory of the particle and the center of gravity of the atom, which is chosen to be the origin, as shown on Fig. 10.1. Let R(t) be the position of the particle at time t and r = (x, y, z) the coordinates of the electron of the atom. The impact parameter is b and the notation is speciﬁed in Fig. 10.1. The time at which the particle passes nearest to the atom, i.e. x = b, y = 0 is denoted t = 0. We write En and |n for the energy levels and corresponding eigenstates of the atom in the absence of an external perturbation.

10.1 Energy Absorbed by One Atom 10.1.1. Write the expression for the time-dependent perturbing potential Vˆ (t) due to the presence of the charged particle.

88

10 Energy Loss of Ions in Matter

10.1.2. We assume that the impact parameter b is much larger than the typical atomic size, i.e. b r, so that |R(t)| |r| for all t. Replace Vˆ (t) by its ﬁrst order expansion in |r|/|R| and express the result in terms of the coordinates x and y of the electron, and of b, v and t. 10.1.3. Initially, at time t = −∞, the atom is in a state |i of energy Ei . Using ﬁrst order time-dependent perturbation theory, write the probability amplitude γif to ﬁnd the atom in the ﬁnal state |f of energy Ef after the h and we only charged particle has passed (t = +∞). We set ωf i = (Ef − Ei )/¯ consider the case Ef = Ei . 10.1.4. The calculation of γif involves the Bessel function K0 (z). One has ∞ ∞ cos ωt sin ωt dt = K0 (ωβ) t 2 dt = ω K0 (ωβ) . 2 + t2 )1/2 (β (β + t2 )3/2 0 0 Express γif in terms of K0 and its derivative. behavior of K0 is K0 (z) − ln z for z 1, and K0 (z) The asymptotic 2π/z e−z for z 1. Under what condition on the parameters ωf i , b and v is the transition probability Pif = |γif |2 large? Show that, in that case, one obtains

Pif

2Z1 e2 hbv ¯

2 |f |ˆ x|i|2 .

10.1.5. Give the physical interpretation of the condition derived above. Show that, given the parameters of the atom, the crucial parameter is the eﬀective interaction time, and give a simple explanation of this eﬀect.

10.2 Energy Loss in Matter We assume in the following that the Hamiltonian of the atom is of the form

Fig. 10.1. Deﬁnition of the coordinates.

10.2 Energy Loss in Matter

89

ˆ2 ˆ0 = p + V (ˆ r) . H 2m 10.2.1. Thomas–Reiche–Kuhn Sum Rule. ˆ 0 ]. (a) Calculate the commutator [ˆ x, H (b) Deduce from this commutator a relation between the matrix elements ˆ 0. i|ˆ x|f and i|ˆ p|f , where |i and |f are eigenstates of H (c) Applying a closure relation to [ˆ x, pˆ] = i¯ h, show that: 2m (Ef − Ei )|f |ˆ x|i|2 = 1 h2 f ¯ for all eigenstates |i of H0 . 10.2.2. Using the Thomas–Reiche–Kuhn sum rule, calculate the expectation value δE of the energy loss of the incident particle when it interacts with the atom. Let E be the energy of the particle before the interaction. Which parameters does the product E δE depend on? 10.2.3. Experimental Application. We are now interested in incident particles which are fully ionized atoms (Z1 = Z, where Z is the atomic number), whose masses are, to a good approximation, proportional to the mass number A = Z + N (where N is the number of neutrons of the isotope). When these ions traverse condensed matter, they interact with many atoms of the medium, and their energy loss implies some averaging over the random impact parameter b. The previous result then takes the form E δE = kZ 2 A , where the constant k depends on the nature of the medium. Semiconductor detectors used for the identiﬁcation of the nuclei in nuclear reactions are based on this result. In the following example, the ions to be identiﬁed are the ﬁnal state products of a reaction induced by 113 MeV nitrogen ions impinging on a target of silver atoms. In Fig. 10.2 each point represents an event, i.e. the energy E and energy loss δE of an ion when it crosses a silicon detector. The reference point corresponds A to the isotope A = 12 of carbon 12 6 C (we use the notation Z N for a nucleus charge Z and mass number A) which loses δE = 30 MeV at an energy E = 50 MeV. (a) Calculate the constant k and the theoretical prediction for the energy loss at 60 and 70 MeV. Put the corresponding points on the ﬁgure. (b) Assuming the reaction could produce the following isotopes: – boron, Z = 5, A = 10, 11, 12 – carbon, Z = 6, A = 11, 12, 13, 14 – nitrogen Z = 7, A = 13, 14, 15, 16,

90

10 Energy Loss of Ions in Matter 10

20

30

40

50

60

70

80

90

100

E(MeV)

10

20

30 12 6

C

40 14 7

N(113MeV)→Ag

50

δE(MeV) Fig. 10.2. Energy loss δE versus energy E through a silicon detector, of the ﬁnal products of a reaction corresponding to 113 MeV nitrogen ions impinging on a target of silver atoms

what nuclei are eﬀectively produced in the reaction? Justify your answers by putting the points corresponding to E = 50 MeV and E = 70 MeV on the ﬁgure.

10.3 Solutions Section 10.1: Energy Absorbed by One Atom 10.1.1. The interaction potential between the particle and the atom is the sum of the Coulomb interactions between the particle and the core, and those between the particle and the outer electron: Z1 e2 Z1 e2 − . Vˆ (t) = R(t) |R(t) − rˆ | 10.1.2. For |R| |r|, we have

−1/2 1 r·R 1 = R2 − 2R · r + r2 + . |R − r| R R3

10.3 Solutions

Therefore

91

2

Z1 e Vˆ (t) − 3 rˆ · R(t) . R (t) Since R(t) = (b, vt, 0),we obtain Vˆ (t) −

Z1 e2 (b2 + v 2 t2 )

3/2

(ˆ xb + yˆvt) .

10.1.3. To ﬁrst order in Vˆ , the probability amplitude is 1 +∞ iωf i t ˆ e f |V (t)|i dt . γif = i¯ h −∞ Inserting the value found above for Vˆ (t), we ﬁnd 1 +∞ Z1 e2 eiωf i t (bf |ˆ x|i + vtf |ˆ y |i) dt . γif = − i¯ h −∞ (b2 + v 2 t2 )3/2 10.1.4. One has ∞ 0

cos ωt dt (β 2

+

3/2 t2 )

=−

1 d ω K0 (ωβ) = − K0 (ωβ) . β dβ β

Setting β = b/v, the amplitude γif is γif = i

2Z1 e2 ωf i (K0 (ωf i b/v) f |ˆ y |i − K0 (ωf i b/v) f |ˆ x|i) . hv 2 ¯

The probability Pif = |γif |2 is large if K0 or K0 are also large. This happens for ωf i b/v 1. In this limit, K0 (z) ∼ − ln z and K0 (z) ∼ −1/z, and we obtain

ωf i b 2Z1 e2 ωf i b ln γif = i f |ˆ x|i − f |ˆ y |i . hvb ¯ v v Since |f |ˆ x|i| |f |ˆ y |i|, one can neglect the second term (x ln x 1 for x 1) and we obtain, for ωf i b/v 1,

Pif = |γif | 2

2Z1 e2 hbv ¯

2 |f |ˆ x|i|2 .

10.1.5. The time τ = b/v is the characteristic time during which the interaction is important, as we can see on the above formulas. For t τ , the interaction is negligible. The condition ωf i τ 1 means that the interaction time τ must be much smaller than the Bohr period ∼ 1/ωf i of the atom. The perturbation Vˆ (t) must have a large Fourier component at ω = ωf i if we want the probability Pif to be signiﬁcant (the shorter in time the perturbation, the larger the

92

10 Energy Loss of Ions in Matter

spread of its Fourier transform in frequency). In the opposite limiting case, where the perturbation is inﬁnitely slow, the atom is not excited. This observation provides an alternative way to evaluate the integrals of question 1.3. The only values of t which contribute signiﬁcantly are those for which t is not too large, compared to τ (say |t| 10 τ ). If ωf i τ 1, one can replace eiωf i t by 1 in these integrals; the second integral is then zero for symmetry reasons and the ﬁrst one is easily evaluated, and gives the desired result. Section 10.2: Energy Loss in Matter 10.2.1. Thomas–Reiche–Kuhn Sum Rule. ˆ 0 ] = i¯ hpˆ/m. (a) We ﬁnd [ˆ x, H (b) Taking the matrix element of this commutator between two eigenstates ˆ 0 , we obtain: |i and |f of H i¯ h ˆ 0 ] | i = (Ei − Ef )f | x f | pˆ | i = f | [ˆ x, H ˆ | i . m (c) We now take the matrix element of [ˆ x, pˆ] = i¯ h between i| and |i and we use the closure relation: i¯ h=

i | x ˆ | f f | pˆ | i −

f

i | pˆ | f f | x ˆ | i

f

m (Ei − Ef ) | f | x ˆ | i | i¯ h

2

2m = (Ei − Ef )| f | x ˆ | i | i¯ h

2

=

−

f

m (Ef − Ei ) | i | x ˆ | f | i¯ h

2

f

,

f

which proves the Thomas–Reiche–Kuhn sum rule. 10.2.2. The expectation value δE of the energy transferred to the atom is δE =

f

(Ef − Ei ) Pif =

2Z1 e2 hbv ¯

2

(Ef − Ei ) | f | x ˆ | i |

2

.

f

Making use of the Thomas–Reiche–Kuhn sum rule, we obtain δE =

2Z12 e4 , mb2 v 2

where m is the electron mass. If the ion has mass M , its kinetic energy is E = M v 2 /2, and we therefore obtain a very simple expression:

10.3 Solutions

M E δE = m

Z1 e2 b

93

2 ,

where we see that the product E δE does not depend on the energy of the incident particle, but is proportional to its mass and to the square of its charge. 10.2.3. With the 126 C point, one obtains k = 3.47. We have put the calculated points of the various isotopes on Fig. 10.3. We make the following observations: (a) For boron, the isotopes 10 B and 11 B are produced, but not 12 B. (b) For carbon, 12 C is produced more abundantly than 13 C, 14 C and 11 C. (c) For nitrogen, there is an abundant production of 14 N, a smaller production of 15 N, but practically no 13 N or 16 N. 10

20

30

10

40

50

60

70

80

10 11 5

90

B

100

11 12 13 14

6

E(MeV)

C

20 14 15 7

N

30 12 6

C

40 14 7

N(113MeV)→Ag

50

δE(MeV) Fig. 10.3. Interpretation of the data of Fig. 10.2

94

10 Energy Loss of Ions in Matter

10.4 Comments Ionization of matter has numerous applications, for instance in developing detectors for particle and nuclear physics, or in deﬁning protection regulations against radioactivity. In order to calculate the energy loss of an ion in matter, one must integrate the above results over the impact parameter. In practice, taking everything into account, one ends up with the following formula, due to Hans Bethe and Felix Bloch, for the rate of energy loss per unit length:

2me c2 β 2 4πK 2 Z 2 e4 N dE = (ln (10.1) − β2) − dx me c2 β 2 I(1 − β 2 ) where β = v/c, K is a constant, N is the number density of atoms in the medium and I is the mean excitation energy of the medium (I ∼ 11.5 eV). The cases of protons or heavy ions is of great interest and, in comparatively recent years, it has allowed a major improvement in the medical treatment of tumors in the eyes (proton therapy) and in the brain (ion therapy) . Owing to the factor 1/β 2 , or equivalently 1/v 2 , in (10.1), practically all the energy is deposited in a very localized region near the stopping point. Figure (10.4) shows the comparison between the eﬀect of ion beams and photons. One can see the enormous advantage, from the medical point of view, of heavy ion beams. These permit to attack and destroy tumors in a very accurate and localized manner, as opposed to γ rays which produce damages all around the point of interest. Pioneering work on brain tumor therapy has been developed in Darmstadt at the Heavy ion accelerator facility. Information can be found on the sites

Fig. 10.4. Energy loss of ions (left) and survival rate of cells (right) as a function of the penetration depth. The dashed curve corresponds to the same quantities for photons. We can see the considerable medical advantage to use heavy ion beams. Document from the data of Heavy ion therapy at GSI, Darmstadt, http://www.gsi.de (Courtesy James Rich)

10.4 Comments

95

http://www-aix.gsi.de/ bio/home.html http://www.sgsmp.ch/protsr-f.htm This promising sector of medical applications in rapidly developing at present.

Part II

Quantum Entanglement and Measurement

12 Schr¨ odinger’s Cat

The superposition principle states that if |φa and |φb are two possible states √ of a quantum system, the quantum superposition (|φa + |φb )/ 2 is also an allowed state for this system. This principle is essential in explaining interference phenomena. However, when it is applied to “large” objects, it leads to paradoxical situations where a system can be in a superposition of states which is classically self-contradictory (antinomic). The most famous example is Schr¨odinger’s “cat paradox” where the cat is in a superposition of the “dead” and “alive” states. The purpose of this chapter is to show that such superpositions of macroscopic states are not detectable in practice. They are extremely fragile, and a very weak coupling to the environment suﬃces to destroy the quantum superposition of the two states |φa and |φb .

12.1 The Quasi-Classical States of a Harmonic Oscillator In this chapter, we shall consider high energy excitations of a one-dimensional harmonic oscillator, of mass m and frequency ω. The Hamiltonian is written 2 ˆ = pˆ + 1 mω 2 x ˆ2 . H 2m 2

ˆ by {|n}. The energy of the state |n is En = We denote the eigenstates of H (n + 1/2)¯ hω. ˆ = x mω/¯h, Pˆ = 12.1.1. Preliminaries. We introduce the operators X ˆ √ pˆ/ m¯hω and the annihilation and creation operators 1 ˆ + iPˆ a ˆ= √ X 2

1 ˆ a ˆ† = √ X − iPˆ 2

ˆ =a N ˆ† a ˆ.

110

12 Schr¨ odinger’s Cat

ˆ Pˆ ] = i, [ˆ ˆ = We recall the commutators: [X, a, a ˆ† ] = 1, and the relations: H ˆ ˆ hω(N + 1/2) and N |n = n|n. ¯ (a) Check that if one works with functions of the dimensionless variables X and P , one has ∂ ˆ =i ∂ . X Pˆ = −i ∂X ∂P ˆ (b) Evaluate the commutator [N , a ˆ], and prove that a ˆ|n =

√

n|n − 1

(12.1)

up to a phase factor which we set equal to 1 in what follows. ˆ and Pˆ , calculate (c) Using (12.1) for n = 0 and expressing a ˆ in terms of X the wave function of the ground state ψ0 (X) and its Fourier transform ϕ0 (P ). It is not necessary to normalize the result. 12.1.2. The Quasi-Classical States. The eigenstates of the operator a ˆ are called quasi-classical states, for reasons which we now examine. Consider an arbitrary complex number α. Show that the following state |α = e−|α|

2

/2

αn √ |n n! n

(12.2)

is a normalized eigenstate of a ˆ with eigenvalue α: a ˆ|α = α|α. 12.1.3. Calculate the expectation value of the energy in a quasi-classical state |α. Calculate also the expectation values x and p and the root mean square deviations ∆x and ∆p for this state. Show that one has ∆x ∆p = h ¯ /2. 12.1.4. Following a similar procedure as in question 1.1(c) above, determine the wave function ψα (X) of the quasi-classical state |α, and its Fourier transform ϕα (P ). Again, it is not necessary to normalize the result. 12.1.5. Suppose that at time t = 0, the oscillator is in a quasi-classical state |α0 with α0 = ρeiφ where ρ is a real positive number. (a) Show that at any later time t the oscillator is also in a quasi-classical state which can be written as e−iωt/2 |α(t). Determine the value of α(t) in terms of ρ, φ, ω and t. (b) Evaluate xt and pt . Taking 1.3, and assuming that |α| 1, justify brieﬂy why these states are called “quasi-classical”. 12.1.6. Numerical Example. Consider a simple pendulum of length 1 meter and of mass 1 gram. Assume the state of this pendulum can be described by a quasi-classical state. At time t = 0 the pendulum is at x0 = 1 micrometer from its classical equilibrium position, with zero mean velocity. (a) What is the corresponding value of α(0)? (b) What is the relative uncertainty on its position ∆x/x0 ? (c) What is the value of α(t) after 1/4 period of oscillation?

12.3 Quantum Superposition Versus Statistical Mixture

111

12.2 Construction of a Schr¨ odinger-Cat State During the time interval [0, T ], one adds to the harmonic potential, the coupling ˆ =h W ¯ g (ˆ a† a ˆ)2 . We assume that g is much larger than ω and that ωT 1. Hence, we can make the approximation that, during the interval [0, T ], the Hamiltonian of ˆ . At time t = 0, the system is in a quasi-classical state the system is simply W |ψ(0) = |α. ˆ , and write the expan12.2.1. Show that the states |n are eigenstates of W sion of the state |ψ(T ) at time T on the basis {|n}. 12.2.2. How does |ψ(T ) simplify in the particular cases T = 2π/g and T = π/g? 12.2.3. One now chooses T = π/2g. Show that this gives 1 −iπ/4 |α + eiπ/4 | − α . e |ψ(T ) = √ 2

(12.3)

12.2.4. Suppose α is pure imaginary: α = iρ. (a) Discuss qualitatively the physical properties of the state (12.3). (b) Consider a value of |α| of the same order of magnitude as in 1.6. In what sense can this state be considered a concrete example of the “Schr¨odinger cat” type of state mentioned in the introduction?

12.3 Quantum Superposition Versus Statistical Mixture We now study the properties of the state (12.3) in a “macroscopic” √ situation hω. |α| 1. We choose α pure imaginary, α = iρ, and we set p0 = ρ 2m¯ 12.3.1. Consider a quantum system in the state (12.3). Write the (non-normalized) probability distributions for the position and for the momentum of the system. These probability distributions are represented in Fig. 12.1 for α = 5i. Interpret these distributions physically. 12.3.2. A physicist (Alice) prepares N independent systems all in the state (12.3) and measures the momentum of each of these systems. The measuring apparatus has a resolution δp such that: √ m¯hω δp p0 . For N 1, draw qualitatively the histogram of the results of the N measurements.

112

12 Schr¨ odinger’s Cat

Fig. 12.1. Probability distributions for the position and for the momentum of a system in the state (12.3) for α = 5i. The quantities X and P are the dimensionless variables introduced in the ﬁrst part of the problem. The vertical scale is arbitrary

12.3.3. The state (12.3) represents the quantum superposition of two states which are macroscopically diﬀerent, and therefore leads to the paradoxical situations mentioned in the introduction. Another physicist (Bob) claims that the measurements done by Alice have not been performed on N quantum systems in the state (12.3), but that Alice is actually dealing with a nonparadoxical “statistical mixture”, that is to say that half of the N systems are in the state |α and the other half in the state | − α. Assuming this is true, does one obtain the same probability distribution as for the previous question for the N momentum measurements? 12.3.4. In order to settle the matter, Alice now measures the position of each of N independent systems, all prepared in the state (12.3). Draw the shape of the resulting distribution of events, assuming that the resolution δx of the measuring apparatus is such that: h ¯ 1 . δx |α| mω 12.3.5. Can Bob obtain the same result concerning the N position measurements assuming he is dealing with a statistical mixture? 12.3.6. Considering the numerical value obtained in the case of a simple pendulum in question 1.6, evaluate the resolution δx which is necessary in order to tell the diﬀerence between a set of N systems in the quantum superposition (12.3), and a statistical mixture consisting in N/2 pendulums in the state |α and N/2 pendulums in the state | − α.

12.4 The Fragility of a Quantum Superposition In a realistic physical situation, one must take into account the coupling of the oscillator with its environment, in order to estimate how long one can discriminate between the quantum superposition (12.3) (that is to say the “Schr¨ odinger cat” which is “alive and dead”) and a simple statistical mixture (i.e. a set of cats (systems), half of which are alive, the other half being dead; each cat being either alive or dead.)

12.4 The Fragility of a Quantum Superposition

113

If the oscillator is initially in the quasi-classical state |α0 and if the environment is in a state |χe (0), the wave function of the total system is the product of the individual wave functions, and the state vector of the total system can be written as the (tensor) product of the state vectors of the two subsystems: |Φ(0) = |α0 |χe (0) . The coupling is responsible for the damping of the oscillator’s amplitude. At a later time t, the state vector of the total system becomes: |Φ(t) = |α1 |χe (t) with α1 = α(t)e−γt ; the number α(t) corresponds to the quasi-classical state one would ﬁnd in the absence of damping (question 1.5(a)) and γ is a real positive number. 12.4.1. Using the result 1.3, give the expectation value of the energy of the oscillator at time t, and the energy acquired by the environment when 2γt 1. 12.4.2. For initial states of the “ Schr¨ odinger cat” type for the oscillator, the state vector of the total system is, at t = 0, 1 |Φ(0) = √ e−iπ/4 |α0 + eiπ/4 | − α0 |χe (0) 2 and, at a later time t, 1 iπ/4 | − α1 |χ(−) |Φ(t) = √ e−iπ/4 |α1 |χ(+) e (t) + e e (t) 2 still with α1 = α(t)e−γt . We choose t such that α1 is pure imaginary, with (+) (−) |α1 | 1. |χe (t) and |χe (t) are two normalized states of the environment that are a priori diﬀerent (but not orthogonal). The probability distribution of the oscillator’s position, measured independently of the state of the environment, is then P(x) =

(+)

1 [|ψα1 (x)|2 + |ψ−α1 (x)|2 2 (−) + 2Re(iψα∗ 1 (x)ψ−α1 (x)χ(+) e (t)|χe (t))]. (−)

Setting η = χe (t)|χe (t) with 0 ≤ η ≤ 1 (η is supposed to be real) and using the results of Sect. 3, describe without any calculation, the result of: (a) N independent position measurements, (b) N independent momentum measurements. Which condition on η allows one to distinguish between a quantum superposition and a statistical mixture?

114

12 Schr¨ odinger’s Cat

12.4.3. In a very simple model, the environment is represented by a second oscillator, of same mass and frequency as the ﬁrst one. We assume that this second oscillator is initially in its ground state |χe (0) = |0. If the coupling between the two oscillators is quadratic, we will take for granted that (±)

(±)

• the states |χe (t) are quasi-classical states: |χe (t) = | ± β, • and that, for short times (γt 1): |β|2 = 2γt|α0 |2 . (a) From the expansion (12.2), show that η = β| − β = exp(−2|β|2 ). (b) Using the expression found in question 4.1 for the energy of the ﬁrst oscillator, determine the typical energy transfer between the two oscillators, above which the diﬀerence between a quantum superposition and a statistical mixture becomes unobservable. 12.4.4. Consider again the simple pendulum described above. Assume the damping time is one year (a pendulum in vacuum with reduced friction). Using the result of the previous question, evaluate the time during which a “Schr¨ odinger cat” state can be observed. Comment and conclude.

12.5 Solutions Section 12.1: The Quasi-Classical States of a Harmonic Oscillator 12.1.1. (a) A simple change of variables gives ∂ p ˆ ∂ 1 ¯h ∂ h ¯ = −i = −i Pˆ = √ =√ mω ∂x ∂X m¯hω m¯hω i ∂x √ mω ∂ ∂ ∂ ˆ = mω x X ˆ= i¯ h = i m¯hω =i h ¯ h ¯ ∂p ∂p ∂P ˆ, a (b) We have the usual relations [N ˆ] = [ˆ a† a ˆ, a ˆ] = [ˆ a† , a ˆ]ˆ a = −ˆ a. Consequently: ˆ, a ˆa [N ˆ]|n = −ˆ a|n ⇒ N ˆ|n = (n − 1)ˆ a|n , ˆ corresponding to the eigenvalue n−1. We know and a ˆ|n is an eigenvector of N from the theory of the one-dimensional harmonic oscillator that the energy levels are not degenerated. Therefore we ﬁnd that a ˆ|n = µ|n − 1, where the coeﬃcient µ is determined by calculating the norm of a ˆ|n: √ a† a ˆ|n = n ⇒ µ = n ˆ a|n2 = n|ˆ up to an arbitrary phase. ˆ + iPˆ )|0 = 0. (c) The equation a ˆ|0 = 0 corresponds to (X

∂ In real space: X + ∂X ψ0 (X) = 0 ⇒ ψ0 (X) ∝ exp −X 2 /2 .

∂ ϕ0 (P ) = 0 ⇒ ϕ0 (P ) ∝ exp −P 2 /2 . In momentum space: P + ∂P

12.5 Solutions

115

12.1.2. One can check directly the relation a ˆ|α = α|α: αn αn √ 2 √ a √ ˆ |n = e−|α| /2 n |n − 1 n! n! n n αn 2 √ |n = α|α = αe−|α| /2 n! n

a ˆ|α = e−|α|

2

/2

The calculation of the norm of |α yields: α|α = e−|α|

2

n

|α|2n n!

=1.

12.1.3. The expectation value of the energy is: ˆ ˆ + 1/2|α = h E = α|H|α =h ¯ ωα|N ¯ ω(|α|2 + 1/2) . For x, and p, we use ¯h h ¯ α|ˆ a+a ˆ† |α = (α + α∗ ) x = 2mω 2mω m¯hω mω¯ h ∗ † α|ˆ a−a ˆ |α = i (α − α) p = −i 2 2 h ¯ h ¯ α|(ˆ a+a ˆ† )2 |α − x2 = ((α + α∗ )2 + 1) − x2 . ∆x2 = 2mω 2mω h/2mω, which is independent of α. Therefore ∆x = ¯ Similarly m¯hω m¯hω α|(ˆ a−a ˆ† )2 |α − p2 = − ((α − α∗ )2 − 1) − p2 2 2 Therefore ∆p = m¯hω/2. The Heisenberg inequality becomes in this case an equality ∆x ∆p = h ¯ /2, independently of the value of α. ∆p2 = −

12.1.4. With the X variable, we have

1 ∂ √ X+ ψα (X) = αψα (X) ∂X 2

√ (X − α 2)2 ⇒ ψα (X) = C exp − 2

Similarly, with the P variable,

i ∂ √ P+ ϕα (P ) = αϕα (P ) ∂P 2

√ (P + iα 2)2 ⇒ ϕα (P ) = C exp − . 2

116

12 Schr¨ odinger’s Cat

12.1.5. (a) |ψ(0) = |α0

αn √ 0 e−iEn t/¯h |n n! n αn 2 √ 0 e−inωt |n = e−|α| /2 e−iωt/2 n! n

|ψ(t) = e−|α|

2

/2

= e−iωt/2 |α(t)

with α(t) = α0 e−iωt = ρe−i(ωt−φ)

(b) xt = 2¯h/(mω) ρ cos(ωt − φ) = x0√cos(ωt − φ) pt = − 2m¯ hω ρ sin(ωt − φ) = −p0 sin(ωt − φ)

with

x0 = ρ

2¯h/(mω)

with

√ p0 = ρ 2m¯ hω

These are the equations of motions of a classical oscillator. Using the result 1.3, we obtain ∆p 1 1 ∆x 1 , 1 = = x0 2ρ p0 2ρ The relative uncertainties on the position and on the momentum of the oscillator are quite accurately deﬁned at any time. Hence the name “quasi-classical state”. 12.1.6. (a) The appropriate choice is x0 = x0 and p0 = 0, i.e. φ = 0 g = 3.13 s−1 ⇒ α(0) = 3.9 109 ω = 2πν = (b) ∆x/x0 = 1/(2α(0)) = 1.3 10−10 . (c) After 1/4 period, eiωt = eiπ/2 = i ⇒ α(T /4) = −i 3.9 109 Section 12.2: Construction of a Schr¨ odinger-Cat State ˆ are simply the previous |n, therefore: 12.2.1. The eigenvectors of W ˆ |n = h W ¯ g n2 |n and |ψ(0) = |α

⇒

|ψ(T ) = e−|α|

12.2.2. If T = 2π/g, then e−ign

2

T

2

2

/2

αn 2 √ e−ign T |n n! n

= e−2iπn = 1 and

|ψ(T ) = |α .

.

12.5 Solutions

If T = π/g, then e−ign

2

T

e−ign

2

2

= e−iπn = 1 if n is even, −1 if n is odd, therefore T

= (−1)n

12.2.3. If T = π/2g, then e−ign if n is odd. We can rewrite this relation as e−ign

2

T

=

117

2

T

⇒

|ψ(T ) = | − α π

2

= e−i 2 n = 1 for n even, and e−ign

2

T

= −i

π π 1 1 (1 − i + (1 + i)(−1)n ) = √ (e−i 4 + ei 4 (−1)n ) 2 2

or, equivalently, 1 |ψ(T ) = √ (e−iπ/4 |α + eiπ/4 | − α) . 2 12.2.4. (a) For α = iρ, in the state |α, the oscillator has a zero mean position and a positive velocity. In the state | − α, the oscillator also has a zero mean position, but a negative velocity. The state 12.3 is a quantum superposition of these two situations. (b) If |α| 1, the states |α and | − α are macroscopically diﬀerent (antinomic). The state 12.3 is a quantum superposition of such states. It therefore constitutes a (peaceful) version of Schr¨ odinger’s cat, where we represent “dead” or “alive” cats by simple vectors of Hilbert space. Section 12.3: Quantum Superposition Versus Statistical Mixture 12.3.1. The probability distributions of the position and of the momentum are P (X) ∝ |e−iπ/4 ψα (X) + eiπ/4 ψ−α (X)|2

√ 2 −iπ/4 1 ∝ e exp − (X − iρ 2) 2

2 √ 1 +eiπ/4 exp − (X + iρ 2)2 2 √ 2 π ∝ e−X cos2 Xρ 2 − 4 −iπ/4 iπ/4 P (P ) ∝ |e ϕα (P ) + e ϕ−α (P )|2 √ 2 √ exp(−(P − ρ 2) ) + exp(−(P + ρ 2)2 ) . In the latter equation, we√have used the fact that, for ρ 1, the two Gaussians √ centered at ρ 2 and −ρ 2 have a negligible overlap. 12.3.2. Alice will ﬁnd two peaks, each of which contains roughly half of the events, centered respectively at p0 and −p0 .

118

12 Schr¨ odinger’s Cat

12.3.3. The statistical mixture of Bob leads to the same momentum distribution as that measured by Alice: the N/2 oscillators in the state |α all lead to a mean momentum +p0 , and the N/2 oscillators in the state | − α to −p0 . Up to this point, there is therefore no diﬀerence, and no paradoxical behavior related to the quantum superposition (12.3). 12.3.4. In the X variable, the resolution of the detector satisﬁes δX

1 1 = |α| ρ

Alice therefore has √ a suﬃcient resolution to observe the oscillations of the function cos2 (Xρ 2 − π/4) in the distribution P (X). The shape of the distribution will therefore reproduce the probability law of X drawn on ﬁgure 12.1, i.e. a modulation of period (¯ hπ 2 /(2mα2 ω))1/2 , with a Gaussian envelope. 12.3.5. If Bob performs a position measurement on the N/2 systems in the state |α, he will ﬁnd a Gaussian distribution corresponding to the probability law P (X) ∝ |ψα (X)|2 ∝ exp(−X)2 . He will ﬁnd the same distribution for the N/2 systems in the state | − α. The sum of his results will be a Gaussian distribution, which is quite diﬀerent from the result expected by Alice. The position measurement should, in principle, allow one to discriminate between the quantum superposition and the statistical mixture. 1 h ¯ −26 m. Unfortu12.3.6. The necessary resolution is δx |α| mω ∼ 5 10 nately, it is impossible to attain such a resolution in practice. Section 12.4: The Fragility of a Quantum Superposition 12.4.1. We have E(t) = h ¯ ω(|α0 |2 e−2γt + 1/2): this energy decreases with time. After a time much longer than γ −1 , the oscillator is in its ground state. This dissipation model corresponds to a zero temperature environment. The mean energy acquired by the environment E(0)−E(t) is, for 2γt 1, ∆E(t) 2¯hω|α0 |2 γt. 12.4.2. (a) The probability distribution of the position keeps its Gaussian envelope, but the contrast of the oscillations is reduced by a factor η. (b) The probability distribution of the momentum is given by P(p) =

1 (|ϕα1 (p)|2 + |ϕ−α1 (p)|2 + 2η Re(iϕ∗−α1 (p)ϕα1 (p))) 2

Since the overlap of the two Gaussians ϕα1 (p) and ϕ−α1 (p) is negligible for |α1 | 1, the crossed term, which is proportional to η√ does not contribute hω. signiﬁcantly. One recovers two peaks centered at ±|α1 | 2m¯ The diﬀerence between a quantum superposition and a statistical mixture can be made by position measurements. The quantum superposition leads to a modulation of spatial period (¯ hπ 2 /(2mα2 ω))1/2 with a Gaussian envelope,

12.6 Comments

119

whereas only the Gaussian is observed on a statistical mixture. In order to see this modulation, it must not be too small, say η ≥ 1/10 . 12.4.3. (a) A simple calculation gives β| − β = e−|β|

2

β ∗n (−β)n n

n!

2

2

= e−|β| e−|β| = e−2|β|

2

2

(b) From the previous considerations, we must have e−2|β| ≥ 1/10, i.e. |β| ≤ 1. For times shorter than γ −1 , the energy of the ﬁrst oscillator is E(t) = E(0) − 2γt|α0 |2 ¯hω . The energy of the second oscillator is ¯ ω(|β(t)|2 + 1/2) = h ¯ ω/2 + 2γt|α0 |2 ¯hω . E (t) = h The total energy is conserved; the energy transferred during time t is ∆E(t) = hω = h ¯ ω|β|2 . In order to distinguish between a quantum superposi2γt|α0 |2 ¯ tion and a statistical mixture, we must have ∆E ≤ ¯hω. In other words, if a single energy quantum h ¯ ω is transferred, it becomes problematic to tell the diﬀerence. 12.4.4. With 1/2γ = 1 year = 3 × 107 seconds, the time it takes to reach |β| = 1 is (2γ|α0 |2 )−1 2 × 10−12 seconds!

12.6 Comments Even for a system as well protected from the environment as what we have assumed for the pendulum, the quantum superpositions of macroscopic states are unobservable. After a very short time, all measurements one can make on a system initially prepared in such a state coincide with those made on a statistical mixture. It is therefore not possible, at present, to observe the eﬀects related to the paradoxical character of a macroscopic quantum superposition. However, it is quite possible to observe “mesoscopic” kittens, for systems which have a limited number of degrees of freedom and are well isolated. The ﬁrst attempts concerned SQUIDS (Josephson junctions in superconducting rings), but the results were not conclusive. The idea developed here is oriented towards quantum optics, and has been proposed by Bernard Yurke and David Stoler, Phys. Rev. Lett. 57, p. 13 (1986). The most conclusive results have been obtained at the Ecole Normale Superieure in Paris, on microwave photons (50 GHz) stored in a superconducting cavity (M. Brune, E. Hagley, J. Dreyer, X. Maitre, A. Maali, C. Wunderlich, J.-M. Raimond, and S. Haroche, Phys.

120

12 Schr¨ odinger’s Cat

Rev. Lett. 77, 4887 (1996)). The ﬁeld stored in the cavity is a quasi-perfect harmonic oscillator. The preparation of the kitten (Sect. 2) is accomplished by sending atoms through the cavity. Dissipation (Sect. 4) corresponds to the very weak residual absorption by the walls of the superconducting cavity. One can devise “kittens” made of 5 or 10 photons (i.e. |α|2 = 5 or 10) and one can check precisely the theory, including the decoherence due to dissipation eﬀects.

13 Quantum Cryptography

Cryptography consists in sending a message to a correspondent and in minimizing the risk for this message to be intercepted by an unwanted outsider. The present chapter shows how quantum mechanics can provide a procedure to achieve this goal. We assume here that Alice (A) wants to send Bob (B) some information which may be coded in the binary system, for instance + + − − − + + − ···

(13.1)

We denote the number of bits of this message by n. Alice wants to send this message to Bob only if she has made sure that no “spy” is listening to the communication.

13.1 Preliminaries ˆ = (¯ ˆ where the set Consider a spin 1/2 particle. The spin operator is S h/2)σ σ ˆi , i = x, y, z are the Pauli matrices. We write |σz = +1 and |σz = −1 for h/2. the eigenstates of Sˆz with respective eigenvalues +¯h/2 and −¯ Consider a particle in the state |σz = +1. One measures the component of the spin along an axis u in the (x, z) plane, deﬁned by the unit vector eu = cos θ ez + sin θ ex ,

(13.2)

where ez and ex are the unit vectors along the z and x axes respectively. We recall that the corresponding operator is ¯ ˆ · eu = h (cos θ σ ˆz + sin θ σ ˆx ) . S 2

(13.3)

13.1.1. Show that the possible results of the measurement are +¯h/2 and −¯ h/2.

122

13 Quantum Cryptography

13.1.2. Show that the eigenstates of the observable (13.3) are (up to a multiplicative constant): |σu = +1 = cos φ |σz = +1 + sin φ |σz = −1 |σu = −1 = − sin φ |σz = +1 + cos φ |σz = −1 h/2 and and express φ in terms of θ. Write the probabilities p± u of ﬁnding +¯ −¯ h/2 when measuring the projection of the spin along the u axis. 13.1.3. What are the spin states after measurements that give the results +¯ h/2 and −¯h/2 along u? 13.1.4. Immediately after such a measurement, one measures the z component of the spin. (a) What are the possible results and what are the probabilities of ﬁnding these results in terms of the results found previously along the u axis (observable (13.3)). h/2 as in (b) Show that the probability to recover the same value Sz = +¯ the initial state |σz = +1 is P++ (θ) = (1 + cos2 θ)/2. (c) Assuming now that the initial state is |σz = −1, what is, for the same h/2 sequence of measurements, the probability P−− (θ) to recover Sz = −¯ in the last measurement?

Fig. 13.1. A source emits a pair (a, b) of spin-1/2 particles. Alice measures the component of the spin of a along a direction θa and Bob measures the component of the spin of b along a direction θb

13.2 Correlated Pairs of Spins A source produces a pair (a, b) of spin-1/2 particles (Fig. 13.1), prepared in the state |ψ = φ(r a , r b )|Σ where the spin state of the two particles is 1 |Σ = √ (|σza = +1 ⊗ |σzb = +1 + |σza = −1 ⊗ |σzb = −1) . 2

(13.4)

13.2 Correlated Pairs of Spins

123

Fig. 13.2. A spy, sitting between the source and Bob, measures the component of the b spin along an axis θs

In other words, the spin variables are decoupled from the space variables (r a , r b ). In (13.4), |σua = ± (speciﬁcally u = z) are the eigenstates of the u component of the spin of particle a, and similarly for b. 13.2.1. Show that this state can also be written as: 1 |Σ = √ (|σxa = +1 ⊗ |σxb = +1 + |σxa = −1 ⊗ |σxb = −1) . 2

(13.5)

13.2.2. The pair of particles (a, b) is prepared in the spin state (13.4), (13.5). As the two particles move away from each another, this spin state remains unchanged (unless a measurement is made). (a) Alice ﬁrst measures the spin component of a along an axis ua of angle θa . What are the possible results and the corresponding probabilities in the two cases θa = 0, i.e. the z axis, and θa = π/2, i.e. the x axis? (b) Show that, after Alice’s measurement, the spin state of the two particles depends as follows on the measurement and its result Axis z z x

Result +¯ h/2 +¯ h/2 −¯ h/2

State |σza |σxa |σxa

= +1 ⊗ |σzb = +1 = +1 ⊗ |σxb = +1 = −1 ⊗ |σxb = −1

¿From then on, why can one ignore particle a as far as spin measurements on b are concerned? ˆ = A⊗ ˆ B, ˆ where A ˆ and (We recall that if |ψ = |u⊗|v is a factorized state and C ˆ act respectively on the spaces of |u and |v, then ψ|C|ψ ˆ ˆ ˆ B = u|A|uv| B|v).

13.2.3. After Alice’s measurement, Bob measures the spin of particle b along an axis ub of angle θb . Give the possible results of Bob’s measurement and their probabilities in terms of Alice’s results in the four following conﬁgurations: (a) θa = 0, θb = 0; (b) θa = 0, θb = π/2; (c) θa = π/2, θb = 0;

124

13 Quantum Cryptography 1. Alice and Bob decide along which axes x and z they will make their measurements. 2. Alice, who controls the source S, prepares an ordered sequence of N n pairs of spins in the state (13.4) (n is the number of bits of the message). She sends the b spins to Bob and keeps the a spins. 3. For each spin that they collect, Alice and Bob measure either the x or the z component. Each of them chooses the x or z direction at random with probability p = 1/2. There is no correlation, for a given pair of spins (a, b), between the axis chosen by Alice and the one chosen by Bob. They both register all their results. 4. Bob selects a subset F N of his measurements. He communicates openly to Alice (by cell phone, www, etc.) the axis and the result of the measurement for each event of this subset. In practice F ∼ 0.5. 5. Alice compares, for this subset F N , her axes and her results with those just communicated by Bob. By doing so, she can tell whether or not a spy is present. If a spy is spotted, the procedure stops and a “physical” search for the spy must be undertaken. Otherwise: 6. Alice makes a public announcement that she is convinced not to have been spied upon, and Bob, still openly, communicates his axes of measurements for the remaining spins. However, he does not communicate the corresponding results. 7. . . . Fig. 13.3. The procedure for quantum cryptography

(d) θa = π/2, θb = π/2. In which cases do the measurements on a and b give with certainty the same result? 13.2.4. Consider the situation θa = 0. Suppose that a “spy” sitting between the source and Bob measures the spin of particle b along an axis us of angle θs as sketched in Fig. 13.2. (a) What are, in terms of θs and of Alice’s ﬁndings, the results of the spy’s measurements and their probabilities? (b) After the spy’s measurement, Bob measures the spin of b along the axis deﬁned by θb = 0. What does Bob ﬁnd, and with what probabilities, in terms of the spy’s results? (c) What is the probability P (θs ) that Alice and Bob ﬁnd the same results after the spy’s measurement? (d) What is the expectation value of P (θs ) if the spy chooses θs at random in the interval [0, 2π] with uniform probability? What is this expectation value if the spy chooses only the two values θs = 0 and θs = π/2 each with the same probability p = 1/2?

13.3 The Quantum Cryptography Procedure

125

13.3 The Quantum Cryptography Procedure In order to transmit conﬁdential information, Alice and Bob use the procedure outlined in Fig. 13.3. Comment on this procedure, and answer the following questions. 13.3.1. How can Alice be sure that a spy is present? 13.3.2. What is the probability that an operating spy will escape being detected? Calculate this probability for F N = 200. 13.3.3. Does the spy become more “invisible” if he knows the system of axes (x, z) chosen by Alice and Bob to perform their measurements? 13.3.4. Comment on the two “experiments” whose results are given in Tables 13.1 and 13.2. Show that a spy has certainly listened to communication 2. What is the probability that a spy listened to communication 1, but remained undetected? 13.3.5. Complete the missing item (number 7 in the above procedure), and indicate how Alice can send her message (13.1) to Bob without using any other spin pairs than the N pairs which Bob and her have already analyzed. Using Table 13.3, tell how, in experiment 1, Alice can send to Bob the message (+, −).

Table 13.1. Experiment 1, performed with 12 pairs of spins. Top: set of axes and results obtained by Alice. Bottom: choices of axes and results publicly communicated by Bob A Spin # 1 2 3 4 5 6 7 8 9 10 11 12 A Axis x x z x z z x z z z x x A Result + − + + − − + + + − + − B Spin # 1 2 3 4 5 6 7 8 9 10 11 12 B Axis x x z x x x B Result + − − + + + Table 13.2. Experiment 2, performed with 12 pairs of spins. Top: set of axes and results obtained by Alice. Bottom: choices of axes and results publicly communicated by Bob A Spin # 1 2 3 4 5 6 7 8 9 10 11 12 A Axis x z z z x x z x x z x z A Result + + − + + − + + − − + + B Spin # 1 2 3 4 5 6 7 8 9 10 11 12 B Axis x x x z z z B Result + + − + + −

126

13 Quantum Cryptography

Table 13.3. Choice of axes publicly communicated by Bob in the framework of experiment 1, after Alice has said she is convinced that she is not being spied upon Spin # 2 5 6 8 9 12 Axis x x x z x x

13.4 Solutions Section 13.1: Preliminaries 13.1.1. The spin observable along the u axis is

h cos θ sin θ ˆ · eˆu = ¯ S . 2 sin θ − cos θ ˆ · eˆu , i.e. The possible results of the measurement are the eigenvalues of S ±¯ h/2. 13.1.2. The corresponding eigenvectors are cos(θ/2)|σz = +1 + sin(θ/2)|σz = −1 |σu = +1 = |σu = −1 = − sin(θ/2)|σz = +1 + cos(θ/2)|σz = −1 , therefore φ = θ/2. The probabilities follow directly: 2 2 + 2 − p± u = |σu = ±1|σz = +1| , pu = cos (θ/2) , pu = sin (θ/2) .

13.1.3. The state after a measurement with the result +¯h/2 (or −¯ h/2) is |σu = +1 (or |σu = −1). 13.1.4. (a) If the measurement along u has given +¯h/2, then the probabilities for the second measurement are: h/2) = |σz = ±1|σu = +1|2 p+ z (±¯ with h/2) = cos2 (θ/2) , p+ h/2) = sin2 (θ/2) . p+ z (+¯ z (−¯ If the measurement along u has given −¯ h/2, then h/2) = cos2 (θ/2) , p− h/2) = sin2 (θ/2) . p− z (−¯ z (+¯ h/2 with probabilities: (b) One recovers Sz = +¯ + (i) p+ h/2) = cos4 (θ/2) if the measurement along u has given +¯h/2, u .pz (+¯ − − h/2) = sin4 (θ/2) if the measurement along u has given −¯ h/2. (ii) pu .pz (+¯

13.4 Solutions

127

Altogether, one has P++ = cos4

1 θ θ + sin4 = (1 + cos2 θ) . 2 2 2

(c) The intermediate results are reversed, but the ﬁnal probability is the same 1 P−− = (1 + cos2 θ) . 2 Section 13.2: Correlated Pairs of Spins 13.2.1. The√z and x eigenstates are related by |σx = ±1 = (|σz = +1 ± |σz = −1)/ 2 . If we make the substitution in expression (13.4), we obtain 1 √ (|σza = +1 + |σza = −1) ⊗ (|σzb = +1 + |σzb = −1) 2 2 + (|σza = +1 − |σza = −1) ⊗ (|σzb = +1 − |σzb = −1) , where the crossed terms disappear. More generally, the state under consideration is actually invariant under rotations around the y axis. In an actual experiment, it would be simpler to work with the singlet state √ 1 |0, 0 = √ (|σza = +1 ⊗ |σzb = −1 − |σza = −1 ⊗ |σzb = +1)/ 2 , 2 where Alice and Bob would simply ﬁnd results of opposite signs by measuring along the same axis. 13.2.2. (a) Alice ﬁnds ±¯ h/2 with p = 1/2 in each case. This result is obtained by noticing that the projector on the eigenstate |σza = +1 is Pˆ+a = |σza = +1σza = +1| ⊗ Iˆb and that p(+¯ h/2) = Σ|Pˆ+a |Σ = 1/2, (and similarly for p(−¯ h/2)). (b) This array of results is a consequence of the reduction of the wave packet. If Alice measures along the z axis, we use (13.4); the normalized projections on the eigenstates of Sˆza are |σza = +1 ⊗ |σzb = +1 (Alice’s result: +¯h/2) h/2). A similar formula holds for and |σza = −1 ⊗ |σzb = −1 (Alice’s result: −¯ a measurement along the x axis, because of the invariance property, and its consequence, (13.5). Any measurement on b (a probability, an expectation value) will imply ˆ b where B ˆ b is a projector expectation values of operators of the type Iˆa ⊗ B or a spin operator. Since the states under consideration are factorized, the corresponding expressions for spin measurements on b will be of the type ˆ b (|σ a = +1 ⊗ |σ b = +1) . (σza = +1| ⊗ σzb = +1|)Iˆa ⊗ B z z

128

13 Quantum Cryptography

This reduces to ˆ b |σ b = +1 = σ b = +1|B ˆ b |σ b = +1 σza = +1|σza = +1.σzb = +1|B z z z where the spin state of a is irrelevant. 13.2.3. For the ﬁrst and second conﬁgurations, we can summarize the results as follows: θa

θb

Alice

Bob

Probability

0 0 0 0

0 0 π/2 π/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

+¯h/2 −¯ h/2 ±¯ h/2 ±¯ h/2

p=1 p=1 p± = 1/2 p± = 1/2

The results for θa = π/2, θb = 0 are identical to those of θa = 0, θb = π/2; similarly, the case θa = π/2, θb = π/2 is identical to θa = 0, θb = 0 (one actually recovers the same result for any θa = θb ). In the two cases (a) and (d), where θa = θb , i.e. when they measure along the same axis, Alice and Bob are sure to ﬁnd the same result. 13.2.4. (a) Concerning the ﬁndings of Alice and of the spy, we have: Alice

Spy

Probability

+¯ h/2 +¯ h/2 −¯ h/2 −¯ h/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

cos2 (θs /2) sin2 (θs /2) sin2 (θs /2) cos2 (θs /2)

(b) Concerning the ﬁndings of Bob and of the spy: Spy

Bob

Probability

+¯ h/2 +¯ h/2 −¯ h/2 −¯ h/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

cos2 (θs /2) sin2 (θs /2) sin2 (θs /2) cos2 (θs /2)

(c) The probability that Alice and Bob ﬁnd the same result has actually been calculated in questions 1.4(b,c), we simply have 1 (1 + cos2 θs ) . 2 (d) Amazingly enough, the two expectation values are the same. On one 2π hand, one has 0 P (θs )dθs /(2π) = 3/4. On the other, since P (0) = 1 and P (π/2) = 1/2, on the average p¯ = 3/4 if the values θs = 0 and θs = π/2 are chosen with equal probabilities. P (θS ) =

13.4 Solutions

129

Section 13.3: The Quantum Cryptography Procedure 13.3.1. Necessarily, if θa = θb , the results of Alice and Bob must be the same. If a single measurement done along the same axis θa = θb gives diﬀerent results for Alice and Bob, a spy is certainly operating (at least in an ideal experiment). If θa = θb , on the average half of the results are the same, half have opposite signs. 13.3.2. The only chance for the spy to remain invisible is that Alice and Bob always ﬁnd the same results when they choose the same axis. For each pair of spins, there is a probability 1/2 that they choose the same axis, and there is in this case a probability 1/4 that they do not ﬁnd the same result if a spy is operating (question 2.4(d)). Therefore, for each pair of spins, there is a probability 1/8 that the spy is detected, and a probability 7/8 that the spy remains invisible. This may seem a quite ineﬃcient detection method. However, for a large FN that the spy remains undetected number of events, the probability (7/8) 200 ≈ 2.5 × 10−12 . is very small. For F N = 200 one has (7/8) 13.3.3. Quite surprisingly, as mentioned above, the spy does not gain anything in ﬁnding out which x and z axes Alice and Bob have agreed on in step 1 of the procedure. 13.3.4. Experiment number 2. Measurements 8 and 12, where the axes are the same, give opposite results: rush upon the spy! In experiment number 1, however, measurements 1, 7 and 11 along the x axis do give the same results and are consistent with the assumption that there is no spy around. However, the number N = 3 is quite small in the present case. If a spy is operating, the probability that he remains undetected is ≈ 40%. 13.3.5. Among the (1 − F )N remaining measurements, Alice selects a sequence of events where the axes are the same and which reproduces her message. She communicates openly to Bob the labels of these events, and Bob can (at last!) read the message on his own set of data. In the present case, Alice tells Bob to look at the results # 8 and # 12, where Bob can read (+, −). Comment: This procedure is presently being developed in several industrial research laboratories. In practice, one uses photon pairs with correlated polarizations rather than spin 1/2 particles. See, for instance C. Bennett, G. Brassard, and A. Ekert, Quantum Cryptography, Scientiﬁc American, Vol. 267, p. 26 (October 1992).

14 Direct Observation of Field Quantization

We consider here a two-level atom interacting with a single mode of the electromagnetic ﬁeld. When this mode is treated quantum mechanically, speciﬁc features occur in the atomic dynamics, such as damping and revivals of the Rabi oscillations.

14.1 Quantization of a Mode of the Electromagnetic Field We recall that in classical mechanics, a harmonic oscillator of mass m and frequency ω/2π obeys the equations of motion dx/dt = p/m and dp/dt = Deﬁn−m ω 2 x where x is the position and p the momentum of the oscillator. √ ing the reduced variables X(t) = x(t) mω/¯h and P (t) = p(t)/ ¯hmω, the equations of motion of the oscillator are dP = −ωX , dt

dX = ωP dt

(14.1)

and the total energy U (t) is given by ¯ω 2 h (X (t) + P 2 (t)) . (14.2) 2 14.1.1. Consider a cavity for electromagnetic waves, of volume V . Throughout this chapter, we consider a single mode of the electromagnetic ﬁeld, of the form B(r, t) = uy b(t) cos kz , E(r, t) = ux e(t) sin kz U (t) =

where ux , uy and uz are an orthonormal basis. We recall Maxwell’s equations in vacuum: ∇ · E(r, t) = 0 ∇ · B(r, t) = 0

∂B(r, t) ∂t 1 ∂E(r, t) ∇ ∧ B(r, t) = 2 c ∂t

∇ ∧ E(r, t) = −

132

14 Direct Observation of Field Quantization

and the total energy U (t) of the ﬁeld in the cavity:

0 2 1 2 E (r, t) + U (t) = B (r, t) d3 r with 0 µ0 c2 = 1 . 2 2µ0 V

(14.3)

(a) Express de/dt and db/dt in terms of k, c, e(t), b(t). (b) Express U (t) in terms of V, e(t), b(t), 0 , µ0 . One can take V . sin2 kz d3 r = cos2 kz d3 r = 2 V V (c) Setting ω = ck and introducing the reduced variables χ(t) =

0 V e(t) 2¯ hω

Π(t) =

V b(t) 2µ0 ¯hω

show that the equations for dχ/dt, dΠ/dt and U (t) in terms of χ, Π and ω are formally identical to equations (14.1) and (14.2). 14.1.2. The quantization of the mode of the electromagnetic ﬁeld under consideration is performed in the same way as that of an ordinary harmonic oscillator. One associates to the physical quantities χ and Π, Hermitian operators ˆ which satisfy the commutation relation χ ˆ and Π ˆ =i. [χ, ˆ Π] The Hamiltonian of the ﬁeld in the cavity is ¯ω 2 ˆ2 . ˆC = h χ ˆ +Π H 2 The energy of the ﬁeld is quantized: En = (n + 1/2) h ¯ ω (n is a non-negative ˆ C with eigenvalue En . integer); one denotes by |n the eigenstate of H The quantum states of the ﬁeld in the cavity are linear combinations of the ¯ ω/2, is called the “vacuum”, and the set {|n}. The state |0, of energy E0 = h hω is called the “n photon state”. A “photon” state |n of energy En = E0 + n¯ corresponds to an elementary excitation of the ﬁeld, of energy h ¯ ω. One introduces√the “creation” and “annihilation” operators of a photon as √ ˆ ˆ ˆ − iΠ)/ 2 and a ˆ = (χ ˆ + iΠ)/ 2 respectively. These operators satisfy a ˆ† = (χ the usual relations: √ a ˆ† |n = n + 1|n + 1 √ a ˆ|n = n|n − 1 if n = 0 and a ˆ|0 = 0 . ˆ =a ˆ C in terms of a ˆ† and a ˆ. The observable N ˆ† a ˆ is called the (a) Express H “number of photons”.

14.2 The Coupling of the Field with an Atom

133

The observables corresponding to the electric and magnetic ﬁelds at a point r are deﬁned as:

hω ¯ ˆ a ˆ+a ˆ† sin kz E(r) = ux 0 V

µ0 ¯ hω † ˆ a ˆ −a ˆ cos kz . B(r) = iuy V The interpretation of the theory in terms of states and observables is the same as in ordinary quantum mechanics. ˆ C |n in an (b) Calculate the expectation values E(r), B(r), and n|H n-photon state. 14.1.3. The following superposition: |α = e

−|α|2 /2

∞ αn √ |n , n! n=0

(14.4)

where α is any complex number, is called a “quasi-classical” state of the ﬁeld. (a) Show that |α is a normalized eigenvector of the annihilation operator a ˆ and give the corresponding eigenvalue. Calculate the expectation value n of the number of photons in that state. (b) Show that if, at time t = 0, the state of the ﬁeld is |ψ(0) = |α, then, at time t, |ψ(t) = e−iωt/2 |(αe−iωt ). (c) Calculate the expectation values E(r)t and B(r)t at time t in a quasi-classical state for which α is real. (d) Check that E(r)t and B(r)t satisfy Maxwell’s equations. (e) Calculate the energy of a classical ﬁeld such that E cl (r, t) = E(r)t ˆ and B cl (r, t) = B(r) t . Compare the result with the expectation value ˆ of HC in the same quasi-classical state. (f) Why do these results justify the name “quasi-classical” state for |α if |α| 1?

14.2 The Coupling of the Field with an Atom Consider an atom at point r 0 in the cavity. The motion of the center of mass of the atom in space is treated classically. Hereafter we restrict ourselves to the two-dimensional subspace of internal atomic states generated by the ground state |f and an excited state |e. The origin of atomic energies is chosen in such a way that the energies of |f and |e are respectively −¯ hωA /2 and +¯ hωA /2 (ωA > 0). In the basis {|f , |e}, one can introduce the operators:

1 0 0 0 0 1 σ ˆz = σ ˆ+ = σ ˆ− = , 0 −1 1 0 0 0

134

14 Direct Observation of Field Quantization

that is to say σ ˆ+ |f = |e and σ ˆ− |e = |f , and the atomic Hamiltonian can ˆ A = − h¯ ωA σ ˆ . be written as: H z 2 The set of orthonormal states {|f, n , |e, n, n ≥ 0} where |f, n ≡ |f ⊗|n and |e, n ≡ |e⊗|n forms a basis of the Hilbert space of the {atom+photons} states. ˆA + H ˆ C , and give the correˆ0 = H 14.2.1. Check that it is an eigenbasis of H sponding eigenvalues. 14.2.2. In the remaining parts of the problem we assume that the frequency of the cavity is exactly tuned to the Bohr frequency of the atom, i.e. ω = ωA . ˆ 0 . Show that, Draw schematically the positions of the ﬁrst 5 energy levels of H ˆ 0 are grouped in degenerate except for the ground state, the eigenstates of H pairs. 14.2.3. The Hamiltonian of the electric dipole coupling between the atom and the ﬁeld can be written as:

ˆ =γ a W ˆσ ˆ+ + a ˆ† σ ˆ− , where γ = −d ¯hω/0 V sin kz0 , and where the electric dipole moment d is determined experimentally. ˆ on the states |f, n and |e, n. (a) Write the action of W ˆ† σ ˆ− correspond? (b) To which physical processes do a ˆσ ˆ+ and a ˆ = H ˆ0 + W ˆ and the corresponding 14.2.4. Determine the eigenstates of H energies. Show that the problem reduces to the diagonalization of a set of 2 × 2 matrices. One hereafter sets: 1 |φ± n = √ (|f, n + 1 ± |e, n) 2 hΩ0 ¯ hω ¯ = γ = −d sin kz0 2 0 V

√ Ω n = Ω0 n + 1 .

± The energies corresponding to the eigenstates |φ± n are denoted En .

14.3 Interaction of the Atom with an “Empty” Cavity In the following, one assumes that the atom crosses the cavity along a line where sin kz0 = 1. An atom in the excited state |e is sent into the cavity prepared in the vacuum state |0. At time t = 0 , when the atom enters the cavity, the state of the system is |e, n = 0. 14.3.1. What is the state of the system at a later time t?

14.4 Interaction of an Atom

with a Quasi-Classical State

135

14.3.2. What is the probability Pf (T ) to ﬁnd the atom in the state f at time T when the atom leaves the cavity? Show that Pf (T ) is a periodic function of T (T is varied by changing the velocity of the atom). 14.3.3. The experiment has been performed on rubidium atoms for a couple of states (f, e) such that d = 1.1 × 10−26 C.m and ω/2π = 5.0 × 1010 Hz. The volume of the cavity is 1.87 × 10−6 m3 (we recall that 0 = 1/(36π109 ) S.I.). The curve ∞ Pf (T ), together with the real part of its Fourier transform J(ν) = 0 cos (2πνT ) Pf (T ) dT , are shown in Fig. 14.1. One observes a damped oscillation, the damping being due to imperfections of the experimental setup. How do theory and experiment compare? (We recall that the Fourier transform of a damped sinusoid in time exhibits a peak at the frequency of this sinusoid, whose width is proportional to the inverse of the characteristic damping time.)

1,0

Pf (t)

0,04

(a)

0,8

J(ν)

(b)

0,03

0,6

0,02

0,4 0,01

0,2 0,0

t (µs) 0

20

40

60

80

100

ν (kHz) 0,00

0

20

40

60

80 100 120 140

Fig. 14.1. (a) Probability Pf (T ) of detecting the atom in the ground state after it crosses a cavity containing zero photons; (b) Fourier transform of this probability, as deﬁned in the text

14.4 Interaction of an Atom with a Quasi-Classical State The atom, initially in the state |e, is now sent into a cavity where a quasiclassical state |α of the ﬁeld has been prepared. At time t = 0 the atom enters the cavity and the state of the system is |e ⊗ |α. 14.4.1. Calculate the probability Pf (T, n) to ﬁnd, at time T , the atom in the state |f and the ﬁeld in the state |n + 1, for n ≥ 0. What is the probability to ﬁnd the atom in the state |f and the ﬁeld in the state |0? 14.4.2. Write the probability Pf (T ) to ﬁnd the atom in the state |f , independently of the state of the ﬁeld, as an inﬁnite sum of oscillating functions.

136

14 Direct Observation of Field Quantization

14.4.3. On Fig. 14.2 are plotted an experimental measurement of Pf (T ) and the real part of its Fourier transform J(ν). The cavity used for this measurement is the same as in Fig. 14.1, but the ﬁeld has been prepared in a quasi-classical state before the atom is sent in. (a) Determine the three frequencies ν0 , ν1 , ν2 which contribute most strongly to Pf (T ). (b) Do the ratios ν1 /ν0 and ν2 /ν0 have the expected values? (c) From the values J(ν0 ) and J(ν1 ), determine an approximate value for the mean number of photons |α|2 in the cavity.

Fig. 14.2. (a) Probability Pf (T ) of measuring the atom in the ground state after the atom has passed through a cavity containing a quasi-classical state of the electromagnetic ﬁeld; (b) Fourier transform of this probability

14.5 Large Numbers of Photons: Damping and Revivals Consider a quasi-classical state |α of the ﬁeld corresponding to a large mean number of photons: |α|2 n0 1, where n0 is an integer. In this case, the probability π(n) to ﬁnd n photons can be cast, in good approximation, in the form:

2n 2 |α| 1 (n − n0 )2 √ exp − π(n) = e−|α| . n! 2 n0 2πn0 This Gaussian limit of the Poisson distribution can be obtained by using the √ Stirling formula n! ∼ nn e−n 2πn and expanding ln π(n) in the vicinity of n = n0 . 14.5.1. Show that this probability takes signiﬁcant values only if n lies in a neighborhood δn of n0 . Give the relative value δn/n0 . 14.5.2. For such a quasi-classical state, one tries to evaluate the probability Pf (T ) of detecting the atom in the state f after its interaction with the ﬁeld. In order to do this,

14.6 Solutions

137

• one linearizes the dependence of Ωn on n in the vicinity of n0 : n − n0 Ωn Ωn0 + Ω0 √ , 2 n0 + 1

(14.5)

• one replaces the discrete summation in Pf (T ) by an integral. (a) Show that, under these approximations, Pf (T ) is an oscillating function of T for short times, but that this oscillation is damped away after a characteristic time TD . Give the value of TD . We recall that ∞ 2 2 2 2 1 √ e−(x−x0 ) /2σ cos(αx) dx = e−α σ /2 cos(αx0 ). −∞ σ 2π (b) Does this damping time depend on the mean value of the number of photons n0 ? (c) Give a qualitative explanation for this damping. 14.5.3. If one keeps the expression of Pf (T ) as a discrete sum, an exact numerical calculation shows that one expects a revival of the oscillations of Pf (T ) for certain times TR large compared to TD , as shown in Fig. 14.3. This phenomenon is called quantum revival and it is currently being studied experimentally. Keeping the discrete sum but using the approximation (14.5), can you explain the revival qualitatively? How does the time of the ﬁrst revival depend on n0 ?

Fig. 14.3. Exact theoretical calculation of Pf (T ) for n 25 photons

14.6 Solutions Section 14.1: Quantization of a Mode of the Electromagnetic Field 14.1.1. (a) The pair of Maxwell equations ∇ · E = 0 and ∇ · B = 0 are satisﬁed whatever the values of the functions e(t) and b(t). The equations

138

14 Direct Observation of Field Quantization

∇ ∧ E = −(∂B/∂t) and c2 ∇ ∧ B = −(∂E/∂t) require that: db = −ke(t) . dt

de = c2 kb(t) dt

(b) The electromagnetic energy can be written as:

0 2 1 2 2 2 e (t) sin kz + b (t) cos kz d3 r U (t) = 2 2µ0 V 0 V 2 V 2 = e (t) + b (t) . 4 2µ0 (c) Under the change of functions suggested in the text, we obtain: !

hω 2 ¯ χ˙ = ω Π U (t) = χ (t) + Π 2 (t) . ˙ Π = −ω χ 2 These two equations are formally identical to the equations of motion of a particle in a harmonic oscillator potential. ˆ = i, we deduce that: 14.1.2. (a) From [χ, ˆ Π] 1 ˆ χ ˆ =1. [χ ˆ + iΠ, ˆ − iΠ] 2 √ √ ˆ = i(ˆ In addition, χ ˆ = (ˆ a+a ˆ† )/ 2 and Π a† − a ˆ)/ 2, i.e.:

hω † 1 ˆC = ¯ H a ˆa ˆ +a ˆ† a ˆ =h ¯ω a ˆ† a ˆ+ , 2 2 ˆ+1 . ˆC = h ¯ω N or H 2 [ˆ a, a ˆ† ] =

(b) For an n photon state, we ﬁnd n|ˆ a|n = n|ˆ a† |n = 0, which results in E(r) = 0

B(r) = 0 .

ˆ C with eigenvalue (n + 1/2)¯ The state |n is an eigenstate of H hω, i.e.

1 HC = n + hω . ¯ 2 14.1.3. (a) The action of a ˆ on |α gives a ˆ|α = e−|α|

2

/2

= αe−|α|

2

∞ αn √ √ n|n − 1 n! n=1

/2

∞ n=1

αn−1 (n − 1)!

|n − 1 = α|α .

14.6 Solutions

139

The vector |α is normalized: α|α = e−|α|

2

∞ (α∗ )n αn =1. n! n=0

The expectation value of the number of photons in that state is: ˆ |α = α|ˆ n = α|N a† a ˆ|α = ||ˆ a|α||2 = |α|2 . (b) The time evolution of |ψ(t) is given by ∞ αn √ e−iω(n+1/2)t |n n! n=0

n ∞ αe−iωt 2 √ |n = e−iωt/2 e−|α| /2 n! n=0

|ψ(t) = e−|α|

2

/2

= e−iωt/2 |(αe−iωt ) . (c) The expectation values of the electric and magnetic ﬁelds are ¯hω ux E(r)t = 2α cos ωt sin kz V 0 ¯hωµ0 uy . B(r)t = −2α sin ωt cos kz V (d) These ﬁelds are of the same type as the classical ﬁelds considered at the beginning of the problem, with hω ¯ ¯hωµ0 cos ωt b(t) = −2α sin ωt . e(t) = 2α 0 V V Given the relation 0 µ0 c2 = 1, we verify that e(t) ˙ = c2 kb(t) and b˙ = −ke(t). Therefore the expectation values of the ﬁeld operators satisfy Maxwell’s equations. (e) The energy of the classical ﬁeld can be calculated using the result of ques¯ ωα2 . This “classical” tion 1.1b. Since cos2 ωt + sin2 ωt = 1, we ﬁnd U (t) = h ˆ C is: energy is therefore time-independent. The expectation value of H hω(N + 1/2) = h ¯ ω(α2 + 1/2) . HC = ¯ It is also time independent (Ehrenfest’s theorem). (f ) For |α| much larger than 1, the ratio U (t)/HC is close to 1. More generally, the expectation value of a physical quantity as calculated for a quantum ﬁeld in the state |α, will be close to the value calculated for a classical ﬁeld such that E cl (r, t) = E(r)t and B cl (r, t) = B(r)t .

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14 Direct Observation of Field Quantization

Fig. 14.4. (a) Positions of the ﬁve ﬁrst energy levels of H0 . (b) Positions of the ﬁve ˆ ˆ =H ˆ0 + W ﬁrst energy levels of H

Section 14.2: The Coupling of the Field with an Atom 14.2.1. One checks that

1 hωA ¯ + n+ ¯hω |f, n , 2 2

hωA 1 ˆ 0 |e, n = ¯ + n+ H ¯hω |e, n . 2 2

ˆ 0 |f, n = H

−

14.2.2. For a cavity which resonates at the atom’s frequency, i.e. if ω = ωA , ˆ0 the couple of states |f, n + 1, |e, n are degenerate. The ﬁrst ﬁve levels of H are shown in Fig. 14.4a. Only the ground state |f, 0 of the atom+ﬁeld system is non-degenerate. ˆ on the basis vectors of H0 is given by: 14.2.3. (a) The action of W √ ˆ |f, n = n γ |e, n − 1 if n ≥ 1 W = 0 if n = 0 √ ˆ W |e, n = n + 1 γ |f, n + 1 . The coupling under consideration corresponds to an electric dipole interaction ˆ · E(r), ˆ ˆ is the observable electric dipole moment of of the form −D where D the atom. ˆ couples the two states of each degenerate pair. The term a (b) W ˆσ ˆ+ corresponds to the absorption of a photon by the atom, which undergoes a transiˆ− corresponds tion from the ground state to the excited state. The term a ˆ† σ to the emission of a photon by the atom, which undergoes a transition from the excited state to the ground state.

14.6 Solutions

141

ˆ is block-diagonal in the eigenbasis of 14.2.4. The operator W ˆ H0 {|f, n, |e, n} . Therefore: ˆ with the eigenvalue 0. ˆ0 + W • The state |f, 0 is an eigenstate of H ˆ • In each eigen-subspace of H0 generated by {|f, n + 1, |e, n} with n ≥ 0, one must diagonalize the 2 × 2 matrix:

(n + 1)¯ hω ¯ hΩn /2 hω hΩn /2 (n + 1)¯ ¯ whose eigenvectors and corresponding eigenvalues are (n ≥ 0): ¯hΩn 2 ¯hΩn . = (n + 1)¯ hω − 2

|φ+ n

corresponding to

En+ = (n + 1)¯ hω +

|φ− n

corresponding to

En−

ˆ0 + W ˆ are shown in Fig. 14.4b. The ﬁrst ﬁve energy levels of H Section 14.3: Interaction of the Atom and an “Empty” Cavity ˆ 14.3.1. We expand the initial state on on the eigenbasis of H:

1 − |ψ(0) = |e, 0 = √ |φ+ 0 − |φ0 . 2 The time evolution of the state vector is therefore given by: + 1 −iE0− t/¯ h − |φ0 |ψ(t) = √ e−iE0 t/¯h |φ+ 0−e 2 −iωt e iΩ0 t/2 − |φ0 . = √ e−iΩ0 t/2 |φ+ 0−e 2 14.3.2. In general, the probability of detecting the atom in the state f , independently of the ﬁeld state, is given by: Pf (T ) =

∞

|f, n|ψ(T )|2 .

n=0

In the particular case of an initially empty cavity,

√only the term n = 1 con− tributes to the sum. Using |f, 1 = |φ+ 0 + |φ0 / 2, we ﬁnd 1 Ω0 T = (1 − cos Ω0 T ) . 2 2 It is indeed a periodic function of T , with angular frequency Ω0 . Pf (T ) = sin2

14.3.3. Experimentally, one measures an oscillation of frequency ν0 = 47 kHz. This result corresponds to the expected value: hω 1 2d ¯ . ν0 = 2π ¯ h 0 V

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14 Direct Observation of Field Quantization

Section 14.4: Interaction of an Atom with a Quasi-Classical State ˆ: ˆ0 + W 14.4.1. Again, we expand the initial state on the eigenbasis of H |ψ(0) = |e ⊗ |α = e−|α| = e−|α|

2

/2

2

/2

∞ αn √ |e, n n! n=0

∞

αn 1 − √ √ |φ+ n − |φn . n! 2 n=0

At time t the state vector is ∞ 2 + − αn 1 −iEn t/¯ h − √ √ e−iEn t/¯h |φ+ |ψ(t) = e−|α| /2 − e |φ . n n n! 2 n=0 We therefore observe that: • the probability to ﬁnd the atom in the state |f and the ﬁeld in the state |0 vanishes for all values of T , • the probability Pf (T, n) can be√obtained from the scalar product of |ψ(t) − and |f, n + 1 = (|φ+ n + |φn ) / 2: 2 2n 2 |α| − 1 −iEn+ t/¯h Pf (T, n) = e−|α| − e−iEn t/¯h e 4 n! 2n 2n 2 |α| 2 |α| 1 Ωn T sin2 = e−|α| (1 − cos Ωn T ) . = e−|α| n! 2 2 n! 14.4.2. The probability Pf (T ) is simply the sum of all probabilities Pf (T, n): Pf (T ) =

∞ n=0

Pf (T, n) =

2 ∞ 1 e−|α| |α|2n − cos Ωn T . 2 2 n=0 n!

14.4.3. (a) The three most prominent peaks of J(ν) occur at the frequencies ν0 = 47 kHz (already found for an empty cavity), ν1 = 65 kHz and ν2 = 81 kHz. (b) The ratios of the are very close to the theoretical √ measured frequencies √ predictions: ν1 /ν0 = 2 and ν2 /ν0 = 3. (c) The ratio J(ν1 )/J(ν0 ) is of the order of 0.9. Assuming the peaks have the same widths, and that these widths are small compared to the splitting ν1 − ν0 , this ratio correponds to the average number of photons |α|2 in the cavity. Actually, the peaks overlap, which makes this determination somewhat inaccurate. If one performs a more sophisticated analysis, taking into account the widths of the peaks, one obtains |α|2 = 0.85 ± 0.04 (see the reference at end of this chapter). Comment: One can also determine |α|2 from the ratio J(ν2 )/J(ν1 ) which should be equal to |α|2 /2. However, the inaccuracy due to the overlap of the peaks is greater than for J(ν1 )/J(ν0 ), owing to the smallness of J(ν2 ).

14.6 Solutions

143

Section 14.5: Large Numbers of Photons: Damping and Revivals 14.5.1. The probability π(n) takes signiﬁcant values only if (n − n0 )2 /(2n0 ) is not much larger than 1, i.e. for integer values of n in a neighborhood of n0 √ of relative extension of the order of 1/ n0 . For n0 1, the distribution π(n) is therefore peaked around n0 . 14.5.2. (a) Consider the result of question 4.2, where we replace Ωn by its approximation (14.5):

∞ 1 1 n − n0 Pf (T ) = − π(n) cos Ωn0 + Ω0 √ T (14.6) 2 2 n=0 2 n0 + 1 We now replace the discrete sum by an integral: 2 1 1 ∞ e−u /(2n0 ) √ Pf (T ) = − 2 2 −∞ 2πn0

u T du . · cos Ωn0 + Ω0 √ 2 n0 + 1 We have extended the lower integration bound from −n0 down to −∞, using √ the fact that the width of the gaussian is n0 n0 . We now develop the expression to be integrated upon:

Ω uT u √0 T = cos (Ωn0 T ) cos cos Ωn0 +Ω0 √ 2 n0 +1 2 n0 + 1

Ω uT √0 − sin (Ωn0 T ) sin . 2 n0 + 1 The sine term does not contribute to the integral (odd function) and we ﬁnd:

1 1 Ω02 T 2 n0 Pf (T ) = − cos (Ωn0 T ) exp − . 2 2 8(n0 + 1) For n0 1, the argument of the exponential simpliﬁes, and we obtain:

1 1 T2 Pf (T ) = − cos (Ωn0 T ) exp − 2 2 2 TD √ with TD = 2 2/Ω0 . (b) In this approximation, the oscillations are damped out in a time TD which is independent of the number of photons n0 . For a given atomic transition (for ﬁxed values of d and ω), this time TD increases like the square root of the volume of the cavity. In the limit of an inﬁnite cavity, i.e. an atom in empty space, this damping time becomes inﬁnite: we recover the usual Rabi oscillation. For a cavity of ﬁnite size, the number of visible oscillations of √ Pf (T ) is roughly νn0 TD ∼ n0 .

144

14 Direct Observation of Field Quantization

(c) The function Pf (T ) is made up of a large number of oscillating functions with similar frequencies. Initially, these diﬀerent functions are in phase, and their sum Pf (T ) exhibits marked oscillations. After a time TD , the various oscillations are no longer in phase with one another and the resulting oscillation of Pf (T ) is damped. One can ﬁnd the damping time by simply estimating the time for which the two frequencies at half width on either side of the maximum of π(n) are out of phase by π: √ √ 1 and n0 ± n0 n0 ± Ωn0 +√n0 TD ∼ Ωn0 −√n0 TD + π 2 Ω0 T D ∼ π .

⇒

14.5.3. Within the approximation (14.5) suggested in the text, equation (14.6) above corresponds to a periodic evolution of period TR =

4π √ n0 + 1 . Ω0

Indeed

Ωn0

n − n0 + Ω0 √ 2 n0 + 1

TR = 4π (n0 + 1) + 2π(n − n0 ) .

We therefore expect that all the oscillating functions which contribute to Pf (T ) will reset in phase at times TR , 2 TR ,. . . The time of the ﬁrst revival, measured in Fig. 14.3, is Ω0 T 64, in excellent agreement with this predic√ tion. Notice that TR ∼ 4 n0 TD , which means that the revival time is always large compared to the damping time. Actually, one can see from the result of Fig. 14.3 that the functions are only partly in phase. This comes from the fact that the numerical calculation has been done with the exact expression of Ωn . In this case, the diﬀerence between two consecutive frequencies Ωn+1 − Ωn is not exactly a constant, contrary to what happens in approximation (14.5); the function Pf (T ) is not really periodic. After a few revivals, one obtains a complicated behavior of Pf (T ), which can be analysed with the techniques developed for the study of chaos.

14.7 Comments The damping phenomenon which we have obtained above is “classical”: one would obtain it within a classical description of the interaction of the ﬁeld and the atom, by considering a ﬁeld whose intensity is not well deﬁned (this would be the analog of a distribution π(n) of the number of photons). On the other hand, the revival comes from the fact that the set of frequencies Ωn is discrete. It is a direct consequence of the quantization of the electromagnetic

14.7 Comments

145

√ √ ﬁeld, in the same way as the occurrence of frequencies ν0 2, ν0 3,. . . in the evolution of Pf (T ) (Sect. 4). The experiments described in this chapter have been performed in Paris, at the Laboratoire Kastler Brossel. The pair of levels (f, e) correspond to very excited levels of rubidium, which explains the large value of the electric dipole moment d. The ﬁeld is conﬁned in a superconducting niobium cavity (Qfactor of ∼ 108 ), cooled down to 0.8 K in order to avoid perturbations to the experiment due to the thermal black body radiation (M. Brune, F. SchmidtKaler, A. Maali, J. Dreyer, E. Hagley, J.-M. Raimond, and S. Haroche, Phys. Rev. Lett. 76, 1800 (1996)).

15 Ideal Quantum Measurement

In 1940, John von Neumann proposed a deﬁnition for an optimal, or “ideal” measurement of a quantum physical quantity. In this chapter, we study a practical example of such a procedure. Our ambition is to measure the excitation number of a harmonic oscillator S by coupling it to another oscillator D whose phase is measured. We recall that, for k integer: N

e

2iπkn N +1

=N +1

for k = p(N + 1), p integer ;

=0

otherwise.

n=0

15.1 Preliminaries: a von Neumann Detector We want to measure a physical quantity A on a quantum system S. We use a detector D devised for such a measurement. There are two stages in the measurement process. First, we let S and D interact. Then, after S and D get separated and do not interact anymore, we read a result on the detector D. We assume that D possesses an orthonormal set of states {|Di } with Di |Dj = δi,j . These states correspond for instance to the set of values which can be read on a digital display. Let |ψ be the state of the system S under consideration, and |D the state of the detector D. Before the measurement, the state of the global system S + D is |Ψi = |ψ ⊗ |D . Let ai and |φi be the eigenvalues and corresponding eigenstates of the ˆ The state |ψ of the system S can be expanded as observable A. αi |φi . (15.1) |ψ = i

148

15 Ideal Quantum Measurement

15.1.1. Using the axioms of quantum mechanics, what are the probabilities p(ai ) to ﬁnd the values ai in a measurement of the quantity A on this state? 15.1.2. After the interaction of S and D, the state of the global system is in general of the form |Ψf = γij |φi ⊗ |Dj (15.2) i,j

We now observe the state of the detector. What is the probability to ﬁnd the detector in the state |Dj ? 15.1.3. After this measurement, what is the state of the global system S +D? 15.1.4. A detector is called ideal if the choice of |D0 and of the coupling S– D leads to coeﬃcients γij which, for any state |ψ of S, verify: |γij | = δi,j |αj |. Justify this designation.

15.2 Phase States of the Harmonic Oscillator ˆ the We consider a harmonic oscillator of angular frequency ω. We note N ˆ ˆ “number” operator, i.e. the Hamiltonian is H = (N + 1/2))¯ hω with eigenhω, N integer ≥ 0. states |N and eigenvalues EN = (N + 12 )¯ Let s be a positive integer. The so-called “phase states” are the family of states deﬁned at each time t by: N =s 1 √ e−iN (ωt+θm ) |N |θm = s + 1 N =0

(15.3)

where θm can take any of the 2s + 1 values θm =

2πm (m = 0, 1, . . . , s) . s+1

(15.4)

15.2.1. Show that the states |θm are orthonormal. 15.2.2. We consider the subspace of states of the harmonic oscillator such that the number of quanta N is bounded from above by some value s. The sets {|N , N = 0, 1, . . . , s} and {|θm , m = 0, 1, . . . , s} are two bases in this subspace. Express the vectors |N in the basis of the phase states. 15.2.3. What is the probability to ﬁnd N quanta in a phase state |θm ? 15.2.4. Calculate the expectation value of the position x ˆ in a phase state, and ﬁnd a justiﬁcation for the name “phase state”. We recall the relation √ √ − 1), where x is the characteristic x ˆ|N = x0 ( N + 1|N + 1 + N |N 0 √ s length of the problem. We set Cs = N =0 N .

15.4 An “Ideal” Measurement

149

15.3 The Interaction between the System and the Detector We want to perform an “ideal” measurement of the number of excitation quanta of a harmonic oscillator. In order to do so, we couple this oscillator S with another oscillator D, which is our detector. Both oscillators have the ˆ S = (ˆ n + 12 )¯ hω are noted same angular frequency ω. The eigenstates of H 1 ˆ ˆ |n, n = 0, 1, . . . , s, those of HD = (N + 2 )¯ hω are noted |N , N = 0, 1...s ˆ are the number operators of S and D. where n ˆ and N We assume that both numbers of quanta n and N are bounded from above by s. The coupling between S and D has the form: ˆ Vˆ = h ¯g n ˆN

(15.5)

This Hamiltonian is realistic. If the two oscillators are two modes of the electromagnetic ﬁeld, it originates from the crossed Kerr eﬀect .

15.3.1. What are the eigenstates and eigenvalues of the total Hamiltonian ˆ =H ˆS + H ˆ D + Vˆ H

?

15.3.2. We assume that the initial state of the global system S + D is factorized as: |Ψ (0) = |ψS ⊗|ψD , with : |ψS = an |n , |ψD = bN |N (15.6) n

N

where we assume that |ψS and |ψD are normalized. We perform a measurement of n ˆ in the state |Ψ (0). What results can one ﬁnd, with what probabilˆ. ities? Answer the same question for a measurement of N 15.3.3. During the time interval [0, t], we couple the two oscillators. The coupling is switched oﬀ at time t. What is the state |Ψ (t) of the system? Is it also a priori factorizable? 15.3.4. Is the probability law for the couple of random variables {n, N } affected by the interaction? Why?

15.4 An “Ideal” Measurement Initially, at time t = 0, the oscillator S is in a state |ψS = oscillator D is prepared in the state |ψD = √

s 1 |N . s + 1 N =0

s n=0

an |n. The

(15.7)

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15 Ideal Quantum Measurement

15.4.1. We switch on the interaction Vˆ during the time interval [0, t]. Express the state |Ψ (t) in terms of the phase states {|θk } of the oscillator D. 15.4.2. We assume the interaction time is t = t0 ≡ 2π/ [g(s + 1)]. Write the state |Ψ (t0 ) of the system. 15.4.3. What is the probability to ﬁnd the value θk in a measurement of the phase of the “detector” oscillator D? 15.4.4. After this measurement has been performed, what is the state of the oscillator S? Describe qualitatively what will happen if one were to choose an interaction time t = t0 . 15.4.5. Comment on the result. In your opinion, why did J. von Neumann consider this as an “ideal” quantum-measurement process?

15.5 Solutions Section 15.1: Preliminaries; a Von Neumann Detector 15.1.1. Since the state of the system is |ψ = i αi |φi , the probability to ﬁnd the value aj in a measurement of A is p(aj ) = |αj |2 . 15.1.2. The state of the global system is γij |φi ⊗ |Dj . |Ψ1 = i,j

The probability pj to ﬁnd the detector in the state |Dj is the sum of the probabilities |γij |2 : |γij |2 , pj = i

since the states |φi are orthogonal. 15.1.3. After this measurement, the state of the global system S + D is, after the principle of wave packet reduction, 1 γij |φi ⊗ |Dj . |Ψ = √ pj i 15.1.4. For an ideal detector, the probability that the detector is in the state |Dj is pj = |αj |2 = p(aj ) and the state of the set system + detector, once we know the state of the detector, is |φj ⊗ |Dj . This is the expected result, given the wave packet reduction principle.

15.5 Solutions

151

Section 15.2: Phase states of the harmonic oscillator 15.2.1. Given the deﬁnition of the phase states, one has: θm |θn = = =

s s 1 iN (ωt+θm ) −iN (ωt+θn ) e e N |N s+1

1 s+1 1 s+1

N =0 N =0 s iN (θm −θn )

e

N =0 s

e2iπN (m−n) / (s+1) = δm,n ,

N =0

where the last equality stands because −s ≤ m − n ≤ s. 15.2.2. The scalar product of a state |N with a phase state is θm |N = (N |θm )∗ = √

1 eiN (ωt+θm ) , s+1

hence the expansion: |N =

s

s 1 θm |N |θm = √ eiN (ωt+θm ) |θm . s + 1 m=0 m=0

15.2.3. Given the deﬁnition of a phase state, the probability to ﬁnd N quanta in a state |θm is 1 p(N, θm ) = |N |θm |2 = . s+1 15.2.4. One obtains θm |ˆ x|θm = 2x0

Cs cos (ωt + θm ) . s+1

The phases of the expectation values of x in two phase states |θm and |θn diﬀer by an integer multiple 2(m − n)π/(s + 1) of the elementary phase 2π/(s + 1). Section 15.3: The Interaction between the System and the Detector 15.3.1. The factorized states |n ⊗ |N are eigenstates of the total Hamiltonian ˆ =H ˆS + H ˆ D + Vˆ = (ˆ ˆ + 1)¯ ˆ , H n+N hω + h ¯g n ˆ⊗N with eigenvalues En,N = (n + N + 1)¯ hω + h ¯ g nN . 15.3.2. The results of measurements and the corresponding probabilities are n = 0, 1, . . . , s, p(n) = |an |2 and N = 0, 1, . . . , s, p(N ) = |bN |2 .

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15 Ideal Quantum Measurement

15.3.3. The state of the system at time t is an bN e−i[(n+N +1)ω+gnN ]t |n ⊗ |N . |Ψ (t) = n

N

In general, it is not factorized. 15.3.4. The probability law for the couple of random variables {n, N } is still p(n, N ) = |an |2 |bN |2 . It is not modiﬁed by the interaction since Vˆ commutes ˆ . The quantities n and N are constants of the motion. with n ˆ and N Section 15.4: An “Ideal” Measurement √ 15.4.1. One has bN = 1/ s + 1, hence 1 an e−i[(n+N +1)ω+gnN )]t |n ⊗ |N . |Ψ (t) = √ s+1 n N Inserting the expansion of the states |N in terms of the phase states, one obtains ei(θm −gnt)N e−i(n+1)ωt an |n ⊗ |θm . |Ψ (t) = s + 1 n m N

15.4.2. If the interaction time is t0 = 2π / [g(s + 1)], this expression reduces to s e−i(n+1)ωt0 an |n ⊗ |θn . (15.8) |Ψ (t0 ) = n=0

15.4.3. The probability to ﬁnd the result θn by measuring the phase of the detecting oscillator D on this state is p(θn ) = |an |2 . 15.4.4. After this measurement, the state of the oscillator S is simply |n (up to an arbitrary phase factor). In the state (15.8), the two systems are perfectly correlated. To a phase state of D there corresponds only one state of number of quanta of S. If one were to choose a time interval diﬀerent from t0 , this correlation would not be perfect. After a measurement of the phase of D, the state of S would be a superposition of states with diﬀerent numbers of quanta. 15.4.5. We see that this procedure, which supposes a well deﬁned interaction time interval between the system and the detector, gives the value of the probability p(n) = |an |2 that S is in a state with n quanta. In addition, after one has read the result θn on the detector, one is sure that S is in the state |n, without having to further interact with it (reduction of the wave packet). In this sense, this procedure does follow exactly the axioms of quantum mechanics on measurement. It is therefore an “ideal” measurement of a quantum physical quantity.

15.6 Comments

153

15.6 Comments One can extend formally this result to other systems than harmonic oscillators. In practice, the case studied here is a simpliﬁcation of the concrete case where the oscillators S and D are modes of the electromagnetic ﬁeld. The Hamiltonian which is eﬀectively encountered in a optically non-linear crystal, comes from the phenomenon called the crossed Kerr eﬀect . In an interferometer, where D is a laser beam split in two parts by a semi-transparent mirror, one can let the signal oscillator S interact with one of the beams. The measurement consists in an interferometric measurement when the two beams of D recombine. This type of experiment has been carried out intensively in recent years. It is also called a “non-destructive” quantum measurement (or QND measurement). One can refer to the article by J.-P. Poizat and P. Grangier, Phys. Rev. Lett. 70, 271 (1993).

16 The Quantum Eraser

This chapter deals with a quantum process where the superposition of two probability amplitudes leads to an interference phenomenon. The two amplitudes can be associated with two quantum paths, as in a double slit interference experiment. We shall ﬁrst show that these interferences disappear if an intermediate measurement gives information about which path has actually been followed. Next, we shall see how interferences can actually reappear if this information is “erased” by a quantum device. We consider a beam of neutrons, which are particles of charge zero and spin 1/2, propagating along the x axis with velocity v. In all what follows, the motion of the neutrons in space is treated classically as a uniform linear motion. Only the evolution of their spin states is treated quantum mechanically.

16.1 Magnetic Resonance The eigenstates of the z component of the neutron spin are noted |n : + and |n : −. A constant uniform magnetic ﬁeld B 0 = B0 uz is applied along the z axis (uz is the unit vector along the z axis). The magnetic moment of the ˆn ˆ n , where γn is the gyromagnetic ratio and S ˆ n = γn S neutron is denoted µ the spin operator of the neutron. 16.1.1. What are the magnetic energy levels of a neutron in the presence of the ﬁeld B 0 ? Express the result in terms of ω0 = −γn B0 . 16.1.2. The neutrons cross a cavity of length L between times t0 and t1 = t0 + L/v. Inside this cavity, in addition to the constant ﬁeld B 0 , a rotating ﬁeld B 1 (t) is applied. The ﬁeld B 1 (t) lies in the (x, y) plane and it has a constant angular frequency ω: B 1 (t) = B1 (cos ωt ux + sin ωt uy ) .

(16.1)

Let |ψn (t) = α+ (t)|n : + + α− (t)|n : − be the neutron spin state at time t, and consider a neutron entering the cavity at time t0 .

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16 The Quantum Eraser

(a) Write the equations of evolution for α± (t) when t0 ≤ t ≤ t1 . We set hereafter ω1 = −γn B1 . (b) Setting α± (t) = β± (t) exp[∓iω(t − t0 )/2], show that the problem reduces to a diﬀerential system with constant coeﬃcients. (c) We assume that we are near the resonance: |ω−ω0 | ω1 , and that terms proportional to (ω − ω0 ) may be neglected in the previous equations. Check that, within this approximation, one has, for t0 ≤ t ≤ t1 , β± (t) = β± (t0 ) cos θ − ie∓iωt0 β∓ (t0 ) sin θ , where θ = ω1 (t − t0 )/2. (d) Show that the spin state at time t1 , when the neutron leaves the cavity, can be written as:

α+ (t0 ) α+ (t1 ) = U (t0 , t1 ) (16.2) α− (t1 ) α− (t0 ) where the matrix U (t0 , t1 ) is

U (t0 , t1 ) =

e−iχ cos φ −ie−iδ sin φ −ieiδ sin φ eiχ cos φ

,

(16.3)

with φ = ω1 (t1 − t0 )/2, χ = ω(t1 − t0 )/2 and δ = ω(t1 + t0 )/2.

16.2 Ramsey Fringes The neutrons are initially in the spin state |n : −. They successively cross two identical cavities of the type described above. This is called Ramsey conﬁguration and it is shown in Fig. 16.1. The same oscillating ﬁeld B 1 (t), given

Fig. 16.1. Ramsey’s conﬁguration; the role of the detecting atom A is speciﬁed in parts 3 and 4

16.2 Ramsey Fringes

157

by 16.1, is applied in both cavities. The modulus B1 of this ﬁeld is adjusted so as to satisfy the condition φ = π/4. The constant ﬁeld B 0 is applied throughout the experimental setup. At the end of this setup, one measures the number of outgoing neutrons which have ﬂipped their spin and are in the ﬁnal state |n : +. This is done for several values of ω in the vicinity of ω = ω0 . 16.2.1. At time t0 , a neutron enters the ﬁrst cavity in the state |n : −. What is its spin state, and what is the probability to ﬁnd it in the state |n : +, when it leaves the cavity? 16.2.2. The same neutron enters the second cavity at time t0 = t1 + T , with T = D/v where D is the distance between the two cavities. Between the two cavities the spin precesses freely around B 0 . What is the spin state of the neutron at time t0 ? 16.2.3. Let t1 be the time when the neutron leaves the second cavity: t1 −t0 = t1 − t0 . Express the quantity δ = ω(t1 + t0 )/2 in terms of ω, t0 , t1 and T . Write the transition matrix U (t0 , t1 ) in the second cavity. 16.2.4. Calculate the probability P+ of detecting the neutron in the state |n : + after the second cavity. Show that it is an oscillating function of (ω0 −ω)T . Explain why this result can be interpreted as an interference process. 16.2.5. In practice, the velocities of the neutrons have some dispersion around the mean value v. This results in a dispersion in the time T to get from one cavity to the other. A typical experimental result giving the intensity of the outgoing beam in the state |n : + as a function of the frequency ν = ω/2π of the rotating ﬁeld B 1 is shown in Fig. 16.2. (a) Explain the shape of this curve by averaging the previous result over the distribution

Fig. 16.2. Intensity of the outgoing beam in the state |n : + as a function of the frequency ω/2π for a neutron beam with some velocity dispersion

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16 The Quantum Eraser

We recall that

∞ −∞

2 2 1 dp(T ) = √ e−(T −T0 ) /2τ dT . τ 2π

cos(ΩT ) dp(T ) = e−Ω

2 2

τ /2

cos(ΩT0 ).

(b) In the above experiment, the value of the magnetic ﬁeld was B0 = 2.57× 10−2 T and the distance D = 1.6 m. Calculate the magnetic moment of the neutron. Evaluate the average velocity v0 = D/T0 and the velocity dispersion δv = v0 τ /T0 of the neutron beam. (c) Which optical interference experiment is the result reminiscent of? 16.2.6. Suppose one inserts between the two cavities of Fig. 16.1 a device which can measure the z component of the neutron spin (the principle of such a detector is presented in the next section). Determine the probability P+,+ of detecting the neutron in the state |n : + between the two cavities and in the state |n : + when it leaves the second cavity, and the probability P−,+ of detecting the neutron in the state |n : − between the cavities and in the state |n : + when it leaves the second cavity. Check that one does not have P+ = P+,+ + P−,+ and comment on this fact.

16.3 Detection of the Neutron Spin State In order to measure the spin of a neutron, one lets it interact during a time ˆ a . Let |a : ± τ with a spin 1/2 atom at rest. The atom’s spin operator is S ˆ be the two eigenstates of the observable Saz . After the interaction between the neutron and the atom, one measures the spin of the atom. Under certain conditions, as we shall see, one can deduce the spin state of the neutron after this measurement. 16.3.1. Spin States of the Atom. Let |a : ±x be the eigenstates of Sˆax and |a : ±y those of Sˆay . Write |a : ±x and |a : ±y in the basis {|a : +, |a : −}. Express |a : ±y in terms of |a : ±x. 16.3.2. We assume that the neutron–atom interaction does not aﬀect the neutron’s trajectory. We represent the interaction between the neutron and the atom by a very simple model. This interaction is assumed to last a ﬁnite time τ during which the neutron–atom interaction Hamiltonian has the form 2A ˆ Snz ⊗ Sˆax , Vˆ = h ¯

(16.4)

where A is a constant. We neglect the action of any external ﬁeld, including B 0 , during the time τ . Explain why Sˆnz and Vˆ commute. Give their common eigenstates and the corresponding eigenvalues.

16.4 A Quantum Eraser

159

16.3.3. We hereafter assume that the interaction time τ is adjusted in such a way that Aτ = π/2 . Suppose the initial state of the system is |ψ(0) = |n : + ⊗ |a : +y . Calculate the ﬁnal state of the system |ψ(τ ). Answer the same question if the initial state is |ψ(0) = |n : − ⊗ |a : +y. 16.3.4. We now suppose that the initial spin state is |ψ(0) = (α+ |n : + + α− |n : −) ⊗ |a : +y. After the neutron–atom interaction described above, one measures the z component Saz of the atom’s spin. (a) What results can one ﬁnd, and with what probabilities? (b) After this measurement, what prediction can one make about the value of the z component of the neutron spin? Is it necessary to let the neutron interact with another measuring apparatus in order to know Snz once the value of Saz is known?

16.4 A Quantum Eraser We have seen above that if one measures the spin state of the neutron between the two cavities, the interference signal disappears. In this section, we will show that it is possible to recover an interference if the information left by the neutron on the detecting atom is “erased” by an appropriate measurement. A neutron, initially in the spin state |n : −, is sent into the two-cavity system. Immediately after the ﬁrst cavity, there is a detecting atom of the type discussed above, prepared in the spin state |a : +y. By assumption, the spin state of the atom evolves only during the time interval τ when it interacts with the neutron. 16.4.1. Write the spin state of the neutron–atom system when the neutron is: (a) just leaving the ﬁrst cavity (time t1 ), before interacting with the atom; (b) just after the interaction with the atom (time t1 + τ ); (c) entering the second cavity (time t0 ); (d) just leaving the second cavity ( time t1 ). 16.4.2. What is the probability to ﬁnd the neutron in the state |n : + at time t1 ? Does this probability reﬂect an interference phenomenon? Interpret the result.

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16 The Quantum Eraser

16.4.3. At time t1 , Bob measures the z component of the neutron spin and Alice measures the y component of the atom’s spin. Assume both measurements give +¯h/2. Show that the corresponding probability reﬂects an interference phenomenon. 16.4.4. Is this result compatible with the conclusion of question 4.2? 16.4.5. In your opinion, which of the following three statements are appropriate, and for what reasons? (a) When Alice performs a measurement on the atom, Bob sees at once an interference appear in the signal he is measuring on the neutron. (b) Knowing the result obtained by Alice on each event, Bob can select a subsample of his own events which displays an interference phenomenon. (c) The experiment corresponds to an interference between two quantum paths for the neutron spin. By restoring the initial state of the atom, the measurement done by Alice erases the information concerning which quantum path is followed by the neutron spin, and allows interferences to reappear. 16.4.6. Alice now measures the component of the atom’s spin along an arbitrary axis deﬁned by the unit vector w. Show that the contrast of the interferences varies proportionally to | sin η|, where cos η = w.uz . Interpret the result.

16.5 Solutions Section 16.1: Magnetic Resonance 16.1.1. The magnetic energy levels are: E± = ∓γn ¯hB0 /2 = ±¯ hω0 /2. 16.1.2. (a) The Hamiltonian is ¯ h H= 2

ω0 ω1 eiωt

ω1 e−iωt −ω0

.

Therefore, the evolution equations are iα˙ + =

ω0 ω1 −iωt α+ + e α− ; 2 2

iα˙ − = −

ω0 ω1 +iωt α− + e α+ . 2 2

(b) With the variables β± (t) = α± (t) exp[±iω(t − t0 )/2], we obtain iβ˙ + =

ω0 − ω ω1 −iωt0 β+ + e β− 2 2

;

iβ˙ − =

ω − ω0 ω1 iωt0 β− + e β+ . 2 2

16.5 Solutions

161

(c) If |ω0 − ω| ω1 , we have, to a good approximation, the diﬀerential system ω1 −iωt0 ω1 iωt0 e e β+ , β− ; iβ˙ − = iβ˙ + = 2 2 whose solution is indeed

β± (t) = β± (t0 ) cos

ω1 (t − t0 ) ω1 (t − t0 ) − i e∓iωt0 β∓ (t0 ) sin . 2 2

(d) Deﬁning φ = ω1 (t1 − t0 )/2 , χ = ω(t1 − t0 )/2 , δ = ω(t1 + t0 )/2, we obtain " # α+ (t1 ) = e−iχ β+ (t1 ) = e−iχ α+ (t0 ) cos φ − iα− (t0 )e−iωt0 sin φ " # α− (t1 ) = eiχ β− (t1 ) = e+iχ α− (t0 ) cos φ − iα+ (t0 )e+iωt0 sin φ , and, therefore,

U=

e−iχ cos φ −i eiδ sin φ

−i e−iδ sin φ eiχ cos φ

.

Section 16.2: Ramsey Fringes 16.2.1. We assume φ = π/4; the initial conditions are: α+ (t0 ) = 0 , α− (t0 ) = 1. At time t1 the state is

1 |ψ(t1 ) = √ −i e−iδ |n : + + eiχ |n : − . 2 √ √ In other words α+ (t1 ) = −ie−iδ / 2, α− (t1 ) = eiχ / 2, and P± = 1/2. 16.2.2. We set T = D/v. The neutron spin precesses freely between the two cavities during time T , and we obtain

1 α+ (t0 ) −ie−iδ e−iω0 T /2 =√ . (16.5) α− (t0 ) eiχ e+iω0 T /2 2 16.2.3. By deﬁnition, t0 = t1 +T and t1 = 2t1 −t0 +T , therefore the transition matrix in the second cavity is

−iχ e cos φ −ie−iδ sin φ U = −ieiδ sin φ eiχ cos φ with φ = φ = ω1 (t1 − t0 )/2, χ = χ = ω(t1 − t0 )/2. Only the parameter δ is changed into δ = ω(t1 + t0 )/2 = ω(3t1 + 2T − t0 )/2 .

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16 The Quantum Eraser

16.2.4. The probability amplitude for detecting the neutron in state |+ after the second cavity is obtained by (i) applying the matrix U to the vector (16.5), (ii) calculating the scalar product of the result with |n : +. We obtain α+ (t1 ) =

1 −i(χ+δ+ω0 T /2) − ie−i(δ −χ−ω0 T /2) . −ie 2

Since δ + χ = ωt1

δ − χ =

ω (3t1 + 2T − t0 − t1 + t0 ) = ω(t1 + T ) , 2

we have α+ (t1 ) = −

i −iω(t1 +T /2) −i(ω0 −ω)T /2 e + ei(ω0 −ω)T /2 . e 2

(16.6)

Therefore, the probability that the neutron spin has ﬂipped in the two-cavity system is (ω − ω0 )T . P+ = |α+ (t1 )|2 = cos2 2 With the approximation of Sect. 1.2c, the probability for a spin ﬂip in a single cavity is independent of ω, and is equal to 1/2. In contrast, the present result for two cavities exhibits a strong modulation of the spin ﬂip probability, between 1 (e.g. for ω = ω0 ) and 0 (e.g. for (ω − ω0 )T = π). This modulation results from an interference process of the two quantum paths corresponding respectively to: • a spin ﬂip in the ﬁrst cavity, and no ﬂip in the second one, • no ﬂip in the ﬁrst cavity and a spin ﬂip in the second one. Each of these paths has a probability 1/2, so that the sum of the probability amplitudes (16.6) is fully modulated. 16.2.5. (a) Since cos2 φ/2 = (1 + cos φ)/2, the averaged probability distribution is cos2

(ω − ω0 )T 2

=

# " 1 1 −(ω−ω0 )2 τ 2 /2 + e cos (ω − ω0 )T0 . 2 2

(16.7)

This form agrees with the observed variation in ω of the experimental signal. The central maximum, which is located at ω/2π = 748.8 kHz corresponds to ω = ω0 . For that value, a constructive interference appears whatever the neutron velocity. The lateral maxima and minima are less peaked, however, since the position of a lateral peak is velocity dependent. The ﬁrst two lateral maxima correspond to (ω − ω0 )T0 ±2π. Their amplitude is reduced, compared to the central peak, by a factor exp(−2π 2 τ 2 /T02 ). (b) The angular frequency ω0 is related to the magnetic moment of the neutron by h ¯ ω0 = 2µn B0 which leads to µn = 9.65 × 10−27 J T−1 . The time T0 can be deduced from the spacing between the central maximum and a

16.5 Solutions

163

lateral one. The ﬁrst lateral maximum occurs at 0.77 kHz from the resonance, hence T0 = 1.3 ms. This corresponds to an average velocity v0 = 1230 m s−1 . The ratio of intensities between the second lateral maximum and the central one is roughly 0.55. This is approximately equal to exp(−8π 2 τ 2 /T02 ), and gives τ /T0 ≈ 0.087, and δv ≈ 110 m s−1 . (c) This experiment can be compared to a Young double slit interference experiment with polychromatic light.The central fringe (corresponding to the peak at ω = ω0 ) remains bright, but the contrast of the interferences decreases rapidly as one departs from the center. In fact, the maxima for some frequencies correspond to minima for others. 16.2.6. The probability P++ is the product of the two probabilities: the probability to ﬁnd the neutron in the state |n : + when it leaves the ﬁrst cavity (p = 1/2) and, knowing that it is in the state |n : +, the probability to ﬁnd it in the same state when it leaves the second cavity (p = 1/2); this gives P+,+ = 1/4. Similarly P−,+ = 1/4. The sum P+,+ + P−,+ = 1/2 does not display any interference, since one has measured in which cavity the neutron spin has ﬂipped. This is very similar to an electron double-slit interference experiment if one measures which slit the electron goes through. Section 16.3: Detection of the Neutron Spin State 16.3.1. By deﬁnition: 1 |a : ±x = √ (|a : + ± |a : −) 2 1 |a : ±y = √ (|a : + ± i|a : −) 2 and these states are related to one another by |a : ±y =

1 ((1 ± i)|a : +x + (1 ∓ i)|a : −x) . 2

16.3.2. The operators Sˆnz and Sˆax commute since they act in two diﬀerent Hilbert spaces; therefore [Sˆnz , Vˆ ] = 0. The common eigenvectors of Sˆnz and Vˆ , and the corresponding eigenvalues are h/2 V = ±A¯ h/2 , |n : + ⊗ |a : ±x Snz = +¯ h/2 V = ∓A¯ h/2 . |n : − ⊗ |a : ±x Snz = −¯ The operators Sˆnz and Vˆ form a complete set of commuting operators as far as spin variables are concerned. 16.3.3. Expanding in terms of the energy eigenstates, one obtains for |ψ(0) = |n : + ⊗ |a : +y : 1 |ψ(τ ) = |n : + ⊗ (1 + i)e−iAτ /2 |a : +x + (1 − i)eiAτ /2 |a : −x , 2

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16 The Quantum Eraser

i.e. for Aτ /2 = π/4: 1 |ψ(τ ) = √ |n : + ⊗ (|a : +x + |a : −x) 2 = |n : + ⊗ |a : + . Similarly, if |ψ(0) = |n : − ⊗ |a : +y, then |ψ(τ ) = i|n : − ⊗ |a : −. Physically, this means that the neutron’s spin state does not change since it is an eigenstate of Vˆ , while the atom’s spin precesses around the x axis with angular frequency A. At time τ = π/(2A), it lies along the z axis. 16.3.4. If the initial state is |ψ(0) = (α+ |n : + + α− |n : −) ⊗ |a : +y, the state after the interaction is |ψ(τ ) = α+ |n : + ⊗ |a : + + iα− |n : − ⊗ |a : − . The measurement of the z component of the atom’s spin gives +¯ h/2, with probability |α+ |2 and state |n : + ⊗ |a : + after the measurement, or −¯ h/2 with probability |α− |2 and state |n : − ⊗ |a : − after the measurement. In both cases, after measuring the z component of the atom’s spin, the neutron spin state is known: it is the same as that of the measured atom. It is not necessary to let the neutron interact with another measuring apparatus in order to know the value of Snz . Section 16.4: A Quantum Eraser 16.4.1. The successive states are: step (a) step (b) step (c)

1 √ −ie−iδ |n : + ⊗ |a : +y + eiχ |n : − ⊗ |a : +y 2

1 −iδ √ −ie |n : + ⊗ |a : + + ieiχ |n : − ⊗ |a : − 2 1 √ − ie−i(δ+ω0 T /2) |n : + ⊗ |a : + 2 +iei(χ+ω0 T /2) |n : − ⊗ |a : − .

Finally, when the neutron leaves the second cavity (step d), the state of the system is: |ψf =

1 − ie−i(δ+ω0 T /2) e−iχ |n : + − ieiδ |n : − ⊗ |a : + 2

+iei(χ+ω0 T /2) −ie−iδ |n : + + eiχ |n : − ⊗ |a : − .

16.4.2. The probability to ﬁnd the neutron in state |+ is the sum of the probabilities for ﬁnding:

16.5 Solutions

165

(a) the neutron in state + and the atom in state +, i.e. the square of the modulus of the coeﬃcient of |n : + ⊗ |a : + (1/4 in the present case), (b) the neutron in state + and the atom in state − (probability 1/4 again). One gets therefore P+ = 1/4 + 1/4 = 1/2: There are no interferences since the quantum path leading in the end to a spin ﬂip of the neutron can be determined from the state of the atom. 16.4.3. One can expand the vectors |a : ± on |a : ±y: 1 |ψf = √ − ie−i(δ+ω0 T /2) e−iχ |n : + − ieiδ |n : − 2 2 ⊗ (|a : +y + |a : −y) i(χ+ω0 T /2) +e −ie−iδ |n : + + eiχ |n : − ⊗ (|a : +y − |a : −y) The probability amplitude that Bob ﬁnds +¯ h/2 along the z axis while Alice ﬁnds +¯ h/2 along the y axis is the coeﬃcient of the term |n : + ⊗ |a : +y in the above expansion. Equivalently, the probability is obtained by projecting the state onto |n : + ⊗ |a : +y, and squaring. One obtains

2 ¯h h ¯ 1 P Snz = , Say = = −ie−i(δ+χ+ω0 T /2) − iei(χ−δ +ω0 T /2) 2 2 8 (ω − ω0 )T 1 , = cos2 2 2 which clearly exhibits a modulation reﬂecting an interference phenomenon. Similarly, one ﬁnds that

(ω − ω0 )T h ¯ h ¯ 1 , P Snz = , Say = − = sin2 2 2 2 2 which is also modulated. 16.4.4. This result is compatible with the result 4.2. Indeed the sum of the two probabilities calculated above is 1/2 as in 4.2. If Bob does not know the result found by Alice, or if Alice does not perform a measurement, which is equivalent from his point of view, Bob sees no interferences. The interferences only arise for the joint probability P (Snz , Say ). 16.4.5. (a) This ﬁrst statement is obviously wrong. As seen in question 4.2, if the atom A is present, Bob no longer sees oscillations (in ω − ω0 ) of the probability for detecting the neutron in the state |+. This probability is equal to 1/2 whatever Alice does. Notice that if the statement were correct, this would imply instantaneous transmission of information from Alice to Bob. By seeing interferences appear, Bob would know immediately that Alice is performing an experiment, even though she may be very far away.

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16 The Quantum Eraser

(b) This second statement is correct. If Alice and Bob put together all their results, and if they select the subsample of events for which Alice ﬁnds +¯ h/2, then the number of events for which Bob also ﬁnds +¯h/2 varies like cos2 ((ω − ω0 )T /2); they recover interferences for this subset of events. In the complementary set, where Alice has found −¯ h/2, the number of Bob’s results giving +¯ h/2 varies like sin2 ((ω − ω0 )T /2). This search for correlations between events occurring in diﬀerent detectors is a common procedure, in particle physics for instance. (c) This third statement, although less precise but more picturesque than the previous one, is nevertheless acceptable. The cos2 ((ω − ω0 )T /2) signal found in Sect. 2 can be interpreted as the interference of the amplitudes corresponding to two quantum paths for the neutron spin which is initially in the state |n : −; either its spin ﬂips in the ﬁrst cavity, or it ﬂips in the second one. If there exists a possibility to determine which quantum path is followed by the system, interferences cannot appear. It is necessary to “erase” this information, which is carried by the atom, in order to observe “some” interferences. After Alice has measured the atom’s spin along the y axis, she has, in some sense “restored” the initial state of the system, and this enables Bob to see some interferences. It is questionable to say that information has been erased: one may feel that, on the contrary, extra information has been acquired. Notice that the statement in the text does not specify in which physical quantity the interferences appear. Notice also that the order of the measurements made by Alice and Bob has no importance, contrary to what this third statement seems to imply. 16.4.6. Alice can measure along the axis w = sin η uy +cos η uz , in the (y, z) plane, for instance. Projecting |ψf onto the eigenstate of Sˆaw with eigenvalue +¯ h/2, i.e. cos(η/2) |a : −,

# a calculation similar to 4.3 leads " |a : + + i sin(η/2) to a probability 1 + sin η cos (ω − ω0 )T /2. If η = 0 or π (measurement along the z axis) there are no interferences. For η = π/2 and 3π/2 or, more generally, if Alice measures in the (x, y) plane, the contrast of the interferences, | sin η|, is maximum.

16.6 Comments Ramsey Fringes with Neutrons. The experimental curve given in the text is taken from J.H. Smith et al., Phys. Rev. 108, 120 (1957). Since then, the technique of Ramsey fringes has been considerably improved. Nowadays one proceeds diﬀerently. One stores neutrons in a “bottle” for a time of the order of 100 s and applies two radiofrequency pulses at the begining and at the end of the storage. The elapsed time between the two pulses is 70 s, compared to 1.3 ms here. This improves enormously the accuracy of the frequency measurement. Such experiments are actually devised to measure the electric dipole moment of the neutron, of fundamental interest in relation to time-reversal

16.6 Comments

167

invariance. They set a very small upper bound on this quantity (K.F. Smith et al., Phys. Lett. 234, 191 (1990)). Non Destructive Quantum Measurements. The structure of the interaction Hamiltonian considered in the text has been chosen in order to provide a simple description of the quantum eraser eﬀect. Realistic examples of nondestructive quantum measurements can be found in J.P. Poizat and P. Grangier, Phys. Rev. Lett. 70, 271 (1993), and S.M. Barnett, Nature, Vol. 362, p. 113, March 1993.

17 A Quantum Thermometer

We study here the measurement of the cyclotron motion of an electron. The particle is conﬁned in a Penning trap and it is coupled to the thermal radiation which causes quantum transitions of the system between various energy levels. In all the chapter, we neglect spin eﬀects. The method and results come from an experiment performed at Harvard University in 1999. We consider an electron of mass M and charge q (q < 0), conﬁned in a Penning trap. This trap consists in the superposition of a uniform magnetic ﬁeld B = Bez (B > 0) and an electric ﬁeld which derives from the potential Φ(r) whose power expansion near the origin is: Φ(r) =

M ωz2 2 2z − x2 − y 2 . 4q

(17.1)

The positive quantity ωz has the dimension of an angular frequency. In all this chapter we set ωc = |q|B/M (ωc is called the cyclotron angular frequency) and we assume that ωz ωc . Useful constants: M = 9.1 10−31 kg ; q = −1.6 10−19 C; h = 6.63 10−34 J s; Boltzmann’s constant kB = 1.38 10−23 J K−1 .

17.1 The Penning Trap in Classical Mechanics We recall that the Lorentz force acting on a charged particle moving in an electromagnetic ﬁeld is F = q(E + v × B). 17.1.1. Check that Φ(r) satisﬁes the Laplace equation ∆Φ = 0. What is the shape of a surface of constant potential Φ(r) =Const? 17.1.2. Show that the classical equation of motion of the electron in the trap is: x ¨ + ωc y˙ −

ωz2 x=0 2

y¨ − ωc x˙ −

ωz2 y=0 2

z¨ + ωz2 z = 0 .

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17 A Quantum Thermometer

17.1.3. What is the type of motion along the z axis? 17.1.4. In order to study the component of the motion in the xy plane, we set α = x + iy. (a) What is the diﬀerential equation satisﬁed by α(t)? (b) We seek a solution of this equation of the form α(t) = α0 eiωt . Show that ω is a solution of the equation: ω 2 − ωc ω +

ωz2 =0. 2

(c) We note ωr and ωl the two roots of this equation with ωr > ωl . Show that: ω2 ωl z . ωr ωc 2ωc 17.1.5. We consider the values B = 5.3 T and ωz /(2π) = 64 MHz. (a) Show that the most general motion of the electron in the Penning trap is the superposition of three harmonic oscillator motions. (b) Calculate the frequencies of these motions. (c) Draw the projection on the xy plane of the classical trajectory of the trapped electron, assuming that αr αl (the positive quantities αr and αl represent the amplitudes of the motions of angular frequencies ωr and ωl ).

17.2 The Penning Trap in Quantum Mechanics ˆ the position and momentum operators of the electron. The We note rˆ and p Hamiltonian of the electron in the Penning trap is, neglecting spin eﬀects: 2 ˆ = 1 (ˆ H p − qA(ˆ r )) + qΦ(ˆ r) , 2M

where the electrostatic potential Φ(r) is given by (17.1). For the magnetic vector potential, we choose the form A(r) = B × r/2. ˆ = 17.2.1. Expand the Hamiltonian and show that it can be written as H ˆ xy + H ˆ z , where H ˆ xy only involves the operators x ˆz H ˆ, yˆ, pˆx and pˆy , while H only involves the operators zˆ and pˆz . ˆ z possess a common eigenbasis? ˆ xy and H Do H 17.2.2. We are now interested in the motion along the z axis. This is called the axial motion. Recall without giving any proof: ˆ z in the ˆ†z which allow to write H (a) the expression of the operators a ˆz and a † † ˆ ˆ ˆ ¯ ωz (Nz + 1/2) with Nz = a ˆz a ˆz and [ˆ az , a ˆz ] = 1; form Hz = h ˆ ˆ (b) the eigenvalues of Nz and Hz .

17.2 The Penning Trap in Quantum Mechanics

171

17.2.3. We now consider the motion in the xy plane under the eﬀect of the ˆ xy . We set Ω = ωc2 − 2ωz2 /2. We introduce the right and left Hamiltonian H ˆl : annihilation operators a ˆr and a i MΩ (ˆ x − iˆ y) + √ (ˆ px − iˆ py ) a ˆr = 4¯ h 4¯ hM Ω i MΩ (ˆ x + iˆ y) + √ (ˆ px + iˆ a ˆl = py ) . 4¯ h 4¯ hM Ω (a) Show that [ˆ ar , a ˆ†r ] = [ˆ al , a ˆ†l ] = 1. (b) Show that any left operator commutes with any right operator, i.e.: ˆl ] = 0 [ˆ ar , a

[ˆ ar , a ˆ†l ] = 0

[ˆ a†r , a ˆl ] = 0

[ˆ a†r , a ˆ†l ] = 0 .

(c) Recall the eigenvalues of n ˆr = a ˆ†r a ˆr and n ˆl = a ˆ†l a ˆl (no proof is required). ˆ l possess a common eigenbasis? Do n ˆ r and n ˆ xy can be written as: (d) Show that the Hamiltonian H ˆ xy = h H ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) , where the angular frequencies ωr and ωl have been introduced in Sect. 17.1. ˆ xy . (e) Deduce from this the eigenvalues of the Hamiltonian H 17.2.4. We note |ψ(t) the state of the system at time t and we deﬁne ar |ψ(t) and al (t) = ψ(t)|ˆ al |ψ(t). Using the Ehrenfest thear (t) = ψ(t)|ˆ orem, calculate dar /dt and dal /dt. Integrate these equations and calculate the expectation value of the electron’s position (x(t), y(t)) in the xy plane. We set ar (0) = ρr e−iφr and al (0) = ρl eiφl , where ρr and ρl are real and positive. Show that the time evolution of the expectation value of the electron position r(t) is similar to the classical evolution found in Sect. 17.1. ˆ corresponding to the eigenvalues 0 17.2.5. We note |φ0 the eigenstate of H ˆ ˆ l and Nz . for each of the operators n ˆr , n (a) Determine the corresponding wave function φ0 (r) (it is not necessary to normalize the result). (b) Using the same numerical values as in question 17.1.5, evaluate the spatial extension of φ0 (r). 17.2.6. The experiment is performed at temperatures T ranging between 0.1 K and 4 K. Compare the characteristic thermal energy kB T to each of the energy quanta of the “cyclotron”, “axial” and “magnetron” motions (associˆz and n ˆ l ). For which of these motions does the ated respectively with n ˆr , N discrete nature of the energy spectrum play an important role?

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17 A Quantum Thermometer

17.3 Coupling of the Cyclotron and Axial Motions We now study a method for detecting the cyclotron motion. This method uses a small coupling between this motion and the axial motion. The coupling is produced by an inhomogeneous magnetic ﬁeld, and it can be described by the additional term in the Hamiltonian: ˆ = M ω2 n ˆ2 . W z ˆr z 2 The experimental conditions are chosen such that = 4 × 10−7 . ˆ +W ˆ using the operators n ˆc = H ˆr , 17.3.1. Write the total Hamiltonian H n ˆ l , pˆz and zˆ. 17.3.2. Show that the excitation numbers of the cyclotron motion (ˆ nr ) and of the magnetron motion (ˆ nl ) are constants of the motion. 17.3.3. Consider the eigensubspace Enr ,nl of n ˆ r and n ˆ l , corresponding to the eigenvalues nr and nl . ˆ c in this subspace. (a) Write the form of H (b) Show that the axial motion is harmonic if the system is prepared in a state belonging to Enr ,nl . Give its frequency in terms of nr and nl . ˆ c inside En ,n . (c) Give the eigenvalues and eigenstates of H r

l

ˆ c can be 17.3.4. Deduce from the previous question that the eigenstates of H labeled by 3 quantum numbers, nr , nl , nz . We write these states as |nr , nl , nz . Give the energy eigenvalues in terms of these quantum numbers and of ωr , ωl , ωz and . 17.3.5. One measures the beat between a highly stable oscillator of frequency ωz /(2π) (delivering a signal proportional to sin(ωz t)) and the current induced in an electric circuit by the axial motion. This latter current is proportional to pz (t). (a) Calculate the time evolution of the expectation values of the position and momentum operators zˆ and pˆz assuming that the state of the electron is restricted to be in the subspace Enr ,nl . We choose the initial conditions ˆ z (t = 0) = z0 and pˆz (t = 0) = 0. (b) To ﬁrst order in , what is the phase diﬀerence ϕ between the detected current and the stable oscillator after a time τ ? Show that the measurement of this phase diﬀerence provides a measurement of the excitation number of the cyclotron motion. 17.3.6. We now assume that the electron is in an arbitrary state cnr ,nl ,nz |nr , nl , nz . |Ψ = nr ,nl ,nz

(a) We measure the phase diﬀerence ϕ on a time interval ranging from t = 0 to t = τ . What are the possible results ϕk of the measurement? Show that

17.4 A Quantum Thermometer

173

this provides a means to determine the excitation number of the cyclotron motion. (b) What is the state of the electron after a measurement giving the result ϕk ? (c) We choose τ = 0.1 s and we assume that the measurement of ϕ is done with an accuracy of π/10. Using the previous values of the physical parameters, show that this accuracy leads to an unambiguous determination of the cyclotron excitation number. (d) After a measurement giving the result ϕk , we let the system evolve for a ˆ c . We then perform length of time T under the action of the Hamiltonian H a new measurement of ϕ. What results do we expect?

17.4 A Quantum Thermometer In practice, the cyclotron motion is in thermal equilibrium with a thermostat at temperature T . We recall that, in that situation, the thermal ﬂuctuations can excite the system in an energy level En with some probability pn . We perform successive measurements of the phase diﬀerence ϕ in the time intervals [0, τ ], [τ, 2τ ], . . ., [(N − 1)τ, N τ ]. The total duration N τ of this series of measurements for a given temperature T is N τ = 3000 seconds, i.e. a total number of results N = 3 × 104 for τ = 0.1 s. By this procedure, one can follow the variation of nr during the time interval N τ , with a time resolution of τ . 17.4.1. Two recordings of this measurement are represented on ﬁgure 17.1 for two diﬀerent temperatures. Comment on these results and explain in particular: (a) to what phenomenon are associated the sudden changes of the signal; (b) what is the fraction of the time during which the electron is in the levels nr = 0, nr = 1, nr = 2, . . . (the accuracy of a usual graduated ruler is suﬃcient). 17.4.2. The probability pn for a system to be in the energy level En is given by the Boltzmann factor pn = N exp(−En /kB T ), where N is a normalization factor. Show that for a one-dimensional harmonic oscillator the ratio pn+1 /pn does not depend on n. 17.4.3. Give an estimate of the two temperatures corresponding to the two recordings of Fig. 17.1. 17.4.4. Fig. 17.2 represents more accurate measurements of the occupation probabilities of the various cyclotron levels for several temperatures of the cryostat which contains the Penning trap. (a) Determine the normalization factor N of the probability law pn for a one-dimensional harmonic oscillator of angular frequency ω in thermal equilibrium with a thermostat at temperature T , and calculate the average excitation number n ¯ . It is convenient to set γ = h ¯ ω/(kB T ).

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17 A Quantum Thermometer

2 1 cyclotron 0 quantum number 2 1 0 0

10

30

20

40

50

time (minutes) Fig. 17.1. Time evolution of the quantum number nr corresponding to the cyclotron motion for two temperatures Ta and Tb

(b) Justify the aspect of the curves of ﬁgure 17.2 and evaluate the corresponding temperatures of the measurements. (c) What is the order of magnitude of the lowest temperature one can measure with such a device? (d) How can one improve the sensitivity of such a “quantum thermometer”?

Occupation probability of the cyclotron level nd

1

0,1

0,01

0

1

2

Cyclotron quantum number nd

Fig. 17.2. Occupation probabilities of the energy states of the cyclotron motion. Each straight line corresponds to a given temperature (the vertical scale is logarithmic)

17.5 Solutions Section 17.1: The Penning Trap in Classical Mechanics 17.1.1. The electric ﬁeld E = −∇Φ is:

17.5 Solutions

⎛ E(r) =

M ωz2 ⎝ 2q

175

⎞

x y⎠ −2z

(17.2)

M ω2

Therefore, one has ∆Φ = −∇ · E = − 2q z (1 + 1 − 2) = 0. The potential satisﬁes Laplace’s equation in the vacuum. The surfaces of constant potential are hyperboloids of axis z. 17.1.2. Using the expression (17.2) for the electric ﬁeld, the equation of motion is: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ x x˙ x ¨ 0 2 M ωz ⎝ y ⎠ + q ⎝ y˙ ⎠ × ⎝ 0 ⎠ , M ⎝ y¨ ⎠ = 2 −2z z˙ z¨ B or, by setting ωc = −qB/M : x ¨ + ωc y˙ −

ωz2 x=0 2

y¨ − ωc x˙ −

ωz2 y=0 2

z¨ + ωz2 z = 0 .

17.1.3. Along the z axis, the motion is harmonic, of angular frequency ωz . 17.1.4. (a) The diﬀerential equation satisﬁed by α(t) is: α ¨ − iωc α˙ −

ωz2 α=0 2

(b) If we search a solution of the form α0 eiωt , we ﬁnd that ω is given by the equation: ω2 −ω 2 + ωc ω − z = 0 2 (c) The roots of this equation are: 1 1 ωr = ωl = ωc + ωc2 − 2ωz2 ωc − ωc2 − 2ωz2 2 2

2 We assume ωz ωc , i.e. ωc2 − 2ωz2 ωc 1 − ωz /ωc2 . The two roots ωr and ωl are given approximately by: ω r ωc

ωl

ωz2 . 2ωc

17.1.5. (a) We have seen previously that the motion along z is harmonic of angular frequency ωz . In order to obtain the motion in the xy plane, we integrate the equation of motion of α: α(t) = αl ei(ωl t+φl ) + αr ei(ωr t+φr ) , where αl and αr are two positive real numbers, and φl and φr are two a priori arbitrary phases. This gives the forms of x(t) and y(t) by taking the real and imaginary parts of this expression:

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17 A Quantum Thermometer

x(t) = αl cos(ωl t + φl ) + αr cos(ωr t + φr ) y(t) = αl sin(ωl t + φl ) + αr sin(ωr t + φr ) . The motion in the xy plane is a superposition of two harmonic motions of angular frequencies ωr and ωl . (b) One ﬁnds ωc /2π = 1.48 × 1011 Hz. The frequencies of the three motions are therefore: ωl /2π 14 kHz

ωz /2π = 64 MHz

ωr /2π 150 GHz

(c) The motion in the xy plane is the superposition of two circular motions, one is of radius αr and has a high frequency (ωr ), and the other is of radius αl and has a much lower frequency (ωl ). Assuming αr αl , this results in a trajectory of the type represented on the following ﬁgure

Section 17.2: The Penning Trap in Quantum Mechanics ˆ z with: ˆ =H ˆ xy + H 17.2.1. The expansion of the Hamiltonian gives H 2 pˆ2 ωc ˆ 2 ˆ xy = pˆx + y + M (ω 2 − 2ω 2 )(ˆ H ˆ2 ) + L z z x +y 2M 2M 8 c 2

2 ˆ z = pˆz + 1 M ω 2 zˆ2 , H z 2M 2 ˆz = where we have introduced the z component of the angular momentum L ˆ xy involves only the operators x ˆ, yˆ and pˆx , pˆy , and since x ˆpˆy − yˆpˆx . Since H ˆ z only involves zˆ, pˆz , the two operators H ˆ xy and H ˆ z commute, and each of H ˆ them commutes with the total Hamiltonian H:

ˆz] = 0 ˆ xy , H [H

ˆ xy , H] ˆ =0 [H

ˆ z , H] ˆ =0. [H

ˆ in the form of a common We can therefore search for an eigenbasis of H ˆz. ˆ xy and H eigenbasis of H ˆ z corresponds to a harmonic motion of an17.2.2. (a) The Hamiltonian H gular frequency ωz . Setting pˆ M ωz ˆz = a a ˆz = x ˆ + i√ and N ˆ†z a ˆz , 2¯ h 2M ¯ hωz ˆz = one easily ﬁnds its spectrum since the Hamiltonian can be written as H ˆ hωz (Nz + 1/2). ¯

17.5 Solutions

177

(b) From the commutation relation [ˆ az , a ˆ†z ] = 1, we deduce that the eigenvalˆ z are therefore ˆ ues of Nz are the non-negative integers nz . The eigenvalues of H of the form ¯hωz (nz + 1/2). 17.2.3. We ﬁrst calculate the generic commutator: M Ω i C= (ˆ x − iη yˆ) + ξ √ (ˆ px − iη pˆy ) , 4¯ h 4¯ hM Ω MΩ i (ˆ x + iη yˆ) − ξ √ (ˆ px + iη pˆy ) 4¯ h 4¯ hM Ω where the four numbers η, ξ, η and ξ are equal to ±1. Using [ˆ x, pˆx ] = [ˆ y , pˆy ] = i¯ h, one ﬁnds: 1 C = (ξ + ξ ) (1 + ηη ) . 4 ˆ†r ] corresponds to η = η = +1 and ξ = ξ = 1, (a) The commutator [ˆ ar , a therefore [ˆ ar , a ˆ†r ] = 1. Similarly, one obtains [ˆ al , a ˆ†l ] = 1 from η = η = −1 and ξ = ξ = 1. (b) The commutator [ˆ ar , a ˆl ] corresponds to ξ = 1 and ξ = −1, hence [ˆ ar , a ˆl ] = 0. Similarly, [ˆ ar , a ˆ†l ] vanishes since it corresponds to η = −η = 1. ˆl ] and [ˆ a†r , a ˆ†l ]) also vanish if we take Therefore the other commutators ([ˆ a†r , a the Hermitian conjugates of the previous commutators. ˆ†r ] = 1 results in the fact that the eigen(c) The commutation relation [ˆ ar , a ˆl. values of n ˆ r are the non-negative integers, and the same holds for n (d) The operators n ˆ r and n ˆ l are expanded as: n ˆ r,l =

ˆz pˆ2x + pˆ2y MΩ 2 1 L + (ˆ x + yˆ2 ) − ± 4¯ hM Ω 4¯ h 2 2¯h

ˆ l ). The sum and diﬀerence where the + (resp.−) sign corresponds to n ˆ r (resp. n of the roots of the equation ω 2 − ωc ω + ωz2 /2 = 0 are: ωr − ωl = ωc2 − 2ωz2 = 2Ω . ωr + ω l = ωc We obtain the anticipated result: ˆ xy = h H ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) ˆ xy can therefore be labeled by the two (non(e) The eigenvectors of H negative) integer quantum numbers nr and nl corresponding to the eigenvalues ˆ l . The corresponding eigenstates are noted |nr , nl . The eigenvalue of n ˆ r and n ˆ xy associated to the vector |nr , nl is ¯hωr (nr + 1/2) − ¯hωl (nl + 1/2). of H 17.2.4. We have: ˆ =h [ˆ ar , H] ¯ ωr [ˆ ar , a ˆ†r a ˆr ] = h ¯ ωr a ˆr .

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17 A Quantum Thermometer

Owing to Ehrenfest theorem we therefore ﬁnd: a˙ r = −iωr ar and a˙ l = +iωl al . The solutions of these two equations are: ar (t) = ar (0) e−iωr t

al (t) = al (0) e+iωl t

The expectation value of the position in the xy plane can be calculated by using: i i MΩ MΩ x+ √ y+ √ px py . i(ar −al ) = ar +al = h ¯ h ¯ hM Ω ¯ ¯hM Ω In other words: h ¯ x(t) = Re (ar (t) + al (t)) MΩ

y(t) =

¯ h Re (iar (t) − ial (t)) . MΩ

Setting ar (0) = ρr e−iφr and al (0) = ρl eiφl , we obtain: h ¯ (ρr cos(ωr t + φr ) + ρl cos(ωl t + φl )) x(t) = MΩ h ¯ (ρr sin(ωr t + φr ) + ρl sin(ωl t + φl )) y(t) = MΩ As in the classical motion, the coordinates x and y are the sums of two sinusoidal functions of angular frequencies ωr and ωl . The x and y components have equal amplitudes and are phase shifted by π/2 with respect to each other. The average motion in the xy plane is therefore the superposition of two uniform circular motions of angular frequencies ωr and ωl . The trajectory is the same as in question 1.5. 17.2.5. (a) The wave function φ0 (r) corresponding to nr = nl = nz = 0 can be written as the product of three functions in the variables x + iy, x − iy and z. This function must satisfy a ˆµ φ0 (r) = 0, with µ = r, l, z. Therefore, ˆl respectively: setting η = + and η = − for a ˆr and a

∂ ∂ MΩ M ωz ∂ − iη + (x − iηy) φ0 (r) = 0 + z φ0 (r) = 0 . ∂x ∂y h ¯ ∂z ¯h By adding and subtracting the two equation for η = ±, we obtain:

∂ MΩ MΩ ∂ + x φ0 (r) = 0 + y φ0 (r) = 0 ∂x h ¯ ∂y ¯h The function φ0 (r) is therefore a product of three gaussian functions in the variables x, y, z: 2 2 2 2 2 φ0 (r) ∝ e−(x +y )/4r0 e−z /4z0

with r0 =

¯ h 2M Ω

and

z0 =

¯h . 2M ωz

17.5 Solutions

179

(b) The probability distribution |φ0 (r)|2 is centered at r = 0. Its extension along the x and y axes is ∆x = ∆y = r0 11 nm. Along the z axis, we have ∆z = z0 380 nm. 17.2.6. For the range of temperatures of interest the ratio kB T /¯hωµ for µ = r, l, z is kB T = 0.014 − 0.6 hωr ¯

kB T = 1.5 × 105 − 6 × 106 hωl ¯

kB T = 30 − 1300 ¯hωz

The discrete nature of the energy spectrum will play a decisive role only for ˆ†r ). Only the ﬁrst three levels of the cyclotron motion (corresponding to a ˆr , a this motion nr = 0, 1, 2, 3 will be occupied signiﬁcantly in this very low temperature domain. For the other components of the motion, of much lower frequencies than the cyclotron motion, one expects that the thermal ﬂuctuations will populate a large number of levels. The “quantum” character of the motions will be hidden under the thermal noise. Section 17.3: Coupling of the Cyclotron and Axial Motions ˆc 17.3.1. In the presence of the axial-cyclotron coupling, the Hamiltonian H is: M ωz2 pˆ2 ˆc = h H (1 + ˆ nr ) zˆ2 . ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) + z + 2M 2 ˆ c . The 17.3.2. It is straightforward to check that n ˆ r and n ˆ l commute with H corresponding physical quantities (i.e. the excitation numbers of the cyclotron and magnetron motions) are therefore constants of the motion. ˆ c involves only 17.3.3. (a) Inside the subspace Enr ,nl , the Hamiltonian H the operators zˆ and pˆz : 2 2 ˆ (nr ,nl ) = pˆz + M ωz (1 + nr ) zˆ2 + En ,n H c r l 2M 2

with Enr ,nl = h ¯ ωr (nr + 1/2) − ¯ hωl (nl + 1/2). (b) If the system is prepared in a state belonging to the subspace Enr ,nl , it will remain in this subspace since nr and nl are constants of the motion. Its ˆ (nr ,nl ) , corresponding to a harmonic motion is described by the Hamiltonian √ Hc oscillator along z, with frequency ωz 1 + nr . ˆ c(nr ,nl ) are ¯hωz (nz + 1/2) √1 + nr + En ,n . The (c) The eigenvalues of H r l corresponding eigenstates are the Hermite functions ψn (Z), where: h. Z = z M ωz (1 + nr )1/2 /¯

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17 A Quantum Thermometer

17.3.4. We can perform the same operation as above inside each subspace ˆ c which we note ˆ l . We thus obtain a basis of eigenstates of H of n ˆ r and n |nr , nl , nz . The eigenvalue corresponding to each eigenstate is: √ Enr ,nl ,nz = h ¯ ωr (nr + 1/2) − ¯ hωl (nl + 1/2) + h ¯ ωz (nz + 1/2) 1 + nr Contrary to the result of Sect. 2, this basis no longer corresponds to factorized functions of the variables x ± iy and z. The coupling between the axial and cyclotron motions induces a correlation between the axial frequency and the state of the cyclotron motion. 17.3.5. (a) If the system is prepared in the subspace Enr ,nl , the axial √ Hamiltonian corresponds to a harmonic oscillator of angular frequency ωz 1 + nr . The Ehrenfest theorem then gives:

√ z(t) = z0 cos ωz t 1 + nr

√ √ pz (t) = −M ωz z0 1 + nr sin ωz t 1 + nr since the evolution equations of the quantum expectation values coincide with the classical equations for a harmonic oscillator. (b) The phase shift accumulated during a time interval τ between the detected current, proportional to pz (t), and the external oscillator is: √ ϕ = ωz τ 1 + nr − ωz τ ωz τ nr . 2 Knowing the time τ , the frequency ωz and the coupling constant , one can deduce the cyclotron excitation number nr . 17.3.6. (a) The possible results of a measurement are the numbers ϕk =

ωz τ k , 2

where k = nr is a non-negative integer. A given experimental result determines unambiguously the excitation number of the cyclotron motion. (b) The measurement postulate of quantum mechanics implies that the state |Ψ of the system after a measurement corresponds to the projection of the state vector before the measurement |Ψ on the eigensubspace corresponding to the measured result: cn(0) ,n ,nz |n(0) |Ψ ∝ r , nl , nz nl ,nz

(0)

r

l

where the integer nr corresponds to the result of the measurement of ϕ. One must further normalize the right hand side in order to obtain the state vector |Ψ .

17.5 Solutions

181

(c) For the values of the parameters given in the text, one ﬁnds φ1 = ωz τ /2 2π × 1.28. The accuracy of π/10 is much smaller than the difference between the phase shifts corresponding to nr and nr + 1, and one can indeed measure unambiguously the excitation numbers nr = 0, 1, 2, . . .. ˆr. (d) A measurement of nr prepares the system in an eigensubspace of n ˆ c , nr is a constant of the motion. Since n ˆ r commutes with the Hamiltonian H Any further measurement of the cyclotron excitation number will therefore give the same result nr , corresponding to the same phase shift ϕk . Of course, this conclusion is no longer true if the system is not fully isolated and interacts with its environment. The coupling with the environment can cause transitions ˆ c , as we shall see in the next section. between diﬀerent eigenstates of H Section 17.4: A Quantum Thermometer 17.4.1. (a) The sudden jumps of the signal are associated with a change of the cyclotron excitation number due to the coupling of the trapped electron with the thermostat. We recall that otherwise nr would be a constant of the motion. (b) In the case of the experimental curve a of Fig. 1, the fractions of time spent in the nl = 0, 1 and 2 levels are 80%, 19% and 1%, respectively. In the case of curve b, the fractions are 97% and 3% for the time spent on nl = 0 and nl = 1, respectively. 17.4.2. For a one-dimensional harmonic oscillator of frequency ω, one ﬁnds: pn+1 e−(n+3/2)¯hω/kB T = −(n+1/2)¯hω/k T = e−¯hω/kB T , B pn e which is independent of n. 17.4.3. For the curves of Fig. 1, we ﬁnd: ¯ ωc / |ln(0.24)| 0.7 ¯hωc . This corre• Curve a: p1 /p0 = 0.24, i.e. kB Ta = h sponds to Ta 5 K. In principle, the determination of the temperature cans also be made using p2 /p1 , but the accuracy is poor compared to that obtained with p1 /p0 . ¯ ωc / |ln(0.03)| 0.29 h ¯ ωc . This corre• Curve b: p1 /p0 = 0.03, i.e. kB Tb = h sponds to Tb 2 K. 17.4.4. (a) The normalization factor is determined from: 1=N

∞

e−(n+1/2)¯hω/kB T .

n=0

This is a geometric series in e−γ : 1=N

e−γ/2 1 − e−γ

→

N = 2 sinh(γ/2) .

182

17 A Quantum Thermometer

The mean excitation number is: ne−nγ d −nγ n ln nPn = −nγ = − e n ¯= dγ ne n n or: n ¯=

1 . eγ − 1

(b) One can see on the above expression that n ¯ is a rapidly increasing function of the temperature. If the temperature is such that γ ∼ 1, i.e. kB T ∼ ¯hωc (or T ∼ 7.1 K for this experiment), the mean excitation number is of the order of (e − 1)−1 ∼ 0.6. Below this temperature, the occupation of the level nl = 0 becomes predominant, as can be seen on the curves of Fig. 17.1. The variation of ln pn as a function of n is linear: ln pn = −n

hωc ¯ + Const kB T

The slope increases as the temperature decreases. The curves of Fig. 2 clearly show this linear variation. They correspond to ratios p1 /p0 equal 0.16 , 0.092 , 0.028 , 0.012 , i.e. temperatures of 1.6 K, 2 K, 3 K and 3.9 K, respectively. (c) In order to measure a temperature with such a device, one must use a statistical sample which is signiﬁcantly populated in the level nl = 1. It is experimentally diﬃcult to go below a probability of 10−2 for the level nr = 1, which corresponds to a temperature T ∼ 1.5 K. (d) In order to improve the sensitivity of this thermometer, one can: • Increase signiﬁcantly the total time of measurement in order to detect occupation probabilities of the level nr = 1 signiﬁcantly less than 10−2 ; • Reduce the value of the magnetic ﬁeld B, in order to reduce the cyclotron frequency ωc , and to increase (for a given temperature) the occupation probability of the level nl = 1 .

The data used in this chapter are extracted from the article of S. Peil and G. Gabrielse, Observing the Quantum Limit of an Electron Cyclotron: QND Measurements of Quantum Jumps between Fock States, Physical Review Letters 83, p. 1287 (1999).

Part III

Complex Systems

19 Properties of a Bose–Einstein Condensate

By cooling down a collection of integer spin atoms to a temperature of less than one micro-Kelvin, one can observe the phenomenon of Bose–Einstein condensation. This results in a situation where a large fraction of the atoms are in the same quantum state. Consequently, the system possesses remarkable coherence properties. We study here the ground state of such an N particle system, hereafter called a condensate. We will show that the nature of the system depends crucially on whether the two-body interactions between the atoms are attractive or repulsive.

19.1 Particle in a Harmonic Trap We consider a particle of mass m placed in a harmonic potential with a frequency ω/2π. The Hamiltonian of the system is ˆ2 1 ˆ = p + mω 2 rˆ 2 , H 2m 2 ˆ = (ˆ where rˆ = (ˆ x, yˆ, zˆ) and p px , pˆy , pˆz ) are respectively the position and momentum operators of the particle. We set a0 = ¯h/(mω). 19.1.1. Recall the energy levels of this system, and its ground state wave function φ0 (r). 19.1.2. We wish to obtain an upper bound on this ground state energy by the variational method. We use a Gaussian trial wave function: 1 ψσ (r) = exp(−r2 /(2σ 2 )) with σ > 0 . (19.1) 3/4 2 (σ π) The values of a relevant set of useful integrals are given below. By varying σ, ﬁnd an upper bound on the ground state energy. Compare the bound with the exact value, and comment on the result.

194

19 Properties of a Bose–Einstein Condensate

Formulas:

|ψσ (r)|2 dx dy dz = 1

|ψσ (r)|4 dx dy dz =

∂ψσ (r) 2 1 ∂x dx dy dz = 2σ 2

σ2 x |ψσ (r)| dx dy dz = 2 2

1 1 (2π)3/2 σ 3

2

19.2 Interactions Between Two Conﬁned Particles We now consider two particles of equal masses m, both placed in the same harmonic potential. We denote the position and momentum operators of the ˆ1, p ˆ2. two particles by rˆ 1 , rˆ 2 and p 19.2.1. In the absence of interactions between the particles, the Hamiltonian of the system is ˆ 21 ˆ2 p 1 1 ˆ = p H + 2 + mω 2 rˆ 21 + mω 2 rˆ 22 . 2m 2m 2 2 (a) What are the energy levels of this Hamiltonian? (b) What is the ground state wave function Φ0 (r 1 , r 2 )? 19.2.2. We now suppose that the two particles interact via a potential v(r 1 − r 2 ). We assume that, on the scale of a0 , this potential is of very short range and that it is peaked around the origin. Therefore, for two functions f (r) and g(r) which vary appreciably only over domains larger than a0 , one has

4π¯ h2 a f (r 1 ) g(r 2 ) v(r 1 − r 2 ) d r1 d r2 m 3

3

f (r) g(r) d3 r .

(19.2)

The quantity a, which is called the scattering length, is a characteristic of the atomic species under consideration. It can be positive (repulsive interaction) or negative (attractive interaction). One can measure for instance that for sodium atoms (isotope 23 Na) a = 3.4 nm, whereas a = −1.5 nm for lithium atoms (isotope 7 Li). (a) Using perturbation theory, calculate to ﬁrst order in a the shift of the ˆ caused by the interaction between the two ground state energy of H atoms. Comment on the sign of this energy shift. (b) Under what condition on a and a0 is this perturbative approach expected to hold?

19.4 Condensates with Repulsive Interactions

195

19.3 Energy of a Bose–Einstein Condensate We now consider N particles conﬁned in the same harmonic trap of angular frequency ω. The particles have pairwise interactions through the potential v(r) deﬁned by (19.2). The Hamiltonian of the system is N N N 2 ˆ p 1 1 i ˆ = H + mω 2 rˆ 2i v(ˆ r i − rˆ j ) . + 2m 2 2 i=1 j=1 i=1 j=i

In order to ﬁnd an (upper) estimate of the ground state energy of the system, we use the variational method with factorized trial wave functions of the type: Ψσ (r 1 , r 2 , . . . , r N ) = ψσ (r 1 ) ψσ (r 2 ) . . . ψσ (r N ) , where ψσ (r) is deﬁned in (19.1). 19.3.1. Calculate the expectation values of the kinetic energy, of the potential energy and of the interaction energy, if the N particle system is in the state |Ψσ : Ek (σ) = Ψσ |

N pˆi 2 i=1

2m

|Ψσ

Ep (σ) = Ψσ |

i=1 N

Eint (σ) = Ψσ |

N 1

2

mω 2 rˆi 2 |Ψσ

N

1 v(ˆ r i − rˆ j )|Ψσ 2 i=1 j=1 j=i

ˆ σ . We set E(σ) = Ψσ |H|Ψ ˜ 19.3.2. We introduce the dimensionless quantities E(σ) = E(σ)/(N ¯hω) and ˜ in terms of σ ˜ . Cast the result in the form σ ˜ = σ/a0 . Give the expression of E

3 1 η 2 ˜ E(σ) = + σ ˜ + 3 4 σ ˜2 σ ˜ and express the quantity η as a function of N , a and a0 . In all what follows, we shall assume that N 1. ˆ 19.3.3. For a = 0, recall the ground state energy of H.

19.4 Condensates with Repulsive Interactions In this part, we assume that the two-body interaction between the atoms is repulsive, i.e. a > 0.

196

19 Properties of a Bose–Einstein Condensate

˜ as a function of σ 19.4.1. Draw qualitatively the value of E ˜ . Discuss the ˜min . variation with η of the position of its minimum E 19.4.2. We consider the case η 1. Show that the contribution of the kinetic ˜ is negligible. In that approximation, calculate an approximate energy to E ˜min . value of E 19.4.3. In this variational calculation, how does the energy of the condensate vary with the number of atoms N ? Compare the prediction with the experimental result shown in Fig. 19.1. 19.4.4. Figure 19.1 has been obtained with a sodium condensate (mass m = 3.8 × 10−26 kg) in a harmonic trap of frequency ω/(2π) = 142 Hz. ¯ ω for this potential. (a) Calculate a0 and h (b) Above which value of N does the approximation η 1 hold? (c) Within the previous model, calculate the value of the sodium atom scattering length that can be inferred from the data of Fig. 19.1. Compare the result with the value obtained in scattering experiments a = 3.4 nm. Is it possible a priori to improve the accuracy of the variational method?

Fig. 19.1. Energy per atom E/N in a sodium condensate, as a function of the number of atoms N in the condensate

19.5 Condensates with Attractive Interactions We now suppose that the scattering length a is negative. ˜ as a function of σ 19.5.1. Draw qualitatively E ˜. 19.5.2. Comment on the approximation (19.2) in the region σ → 0.

19.6 Solutions

197

˜ no 19.5.3. Show that there exists a critical value ηc of |η| above which E longer has a local minimum for a value σ ˜ = 0. Calculate the corresponding size σc as a function of a0 . 19.5.4. In an experiment performed with lithium atoms (m = 1.17 × 10−26 kg), it has been noticed that the number of atoms in the condensate never exceeds 1200 for a trap of frequency ω/(2π) = 145 Hz. How can this result be explained?

19.6 Solutions Section 19.1: Particle in a Harmonic Trap 19.1.1. The Hamiltonian of a three-dimensional harmonic oscillator can be written ˆ =H ˆx + H ˆy + H ˆz , H ˆ i represents a one dimensional harmonic oscillator of same frequency where H ˆ of along the axis i = x, y, z. We therefore use a basis of eigenfunctions of H the form φ(x, y, z) = χnx (x) χny (y) χnz (z), i.e. products of eigenfunctions of ˆy, H ˆ z , where χn (x) is the nth Hermite function. The eigenvalues of H ˆ ˆ x, H H hω, where n = nx + ny + nz is a non-negative can be written as En = (n + 3/2)¯ integer. The ground state wave function, of energy (3/2)¯ hω, corresponds to nx = ny = nz = 0, i.e. 1 φ0 (r) = 2 3/4 exp[−r2 /(2a20 )] . (a0 π) 19.1.2. The trial wave functions ψσ are normalized. In order to obtain an ˆ we must calculate E(σ) = upper bound for the ground-state energy of H, ˆ σ and minimize the result with respect to σ. Using the formulas ψσ |H|ψ given in the text, one obtains pˆ2 1 h2 1 ¯ mω 2 σ 2 2 2 |ψσ = 3 mω | r |ψ ψ =3 ψσ | σ σ 2 2m 2m 2σ 2 2 2 and E(σ) =

3 hω ¯ 4

a20 σ2 + 2 2 σ a0

.

¯ ω. In This quantity is minimum for σ = a0 , and we ﬁnd Emin (σ) = (3/2) h this particular case, the upper bound coincides with the exact result. This is due to the fact that the set of trial wave functions contains the ground state ˆ wave function of H.

198

19 Properties of a Bose–Einstein Condensate

Section 19.2: Interactions between Two Conﬁned Particles ˆ 2 , where H ˆ1 ˆ can be written as H ˆ =H ˆ1 + H 19.2.1. (a) The Hamiltonian H ˆ and H2 are respectively the Hamiltonians of particle 1 and particle 2. A basis ˆ is formed by considering products of eigenfunctions of eigenfunctions of H ˆ 2 (functions of ˆ of H1 (functions of the variable r 1 ) and eigenfunctions of H hω, where n is a the variable r 2 ). The energy eigenvalues are En = (n + 3)¯ non-negative integer. ˆ is: (b) The ground state of H Φ0 (r 1 , r 2 ) = φ0 (r 1 ) φ0 (r 2 ) =

1 a30 π 3/2

" # exp −(r12 + r22 )/(2a20 ) .

ˆ is non-degenerate, its shift to ﬁrst 19.2.2. (a) Since the ground state of H order in a can be written as ∆E = Φ0 |˜ v |Φ0 = |Φ0 (r 1 , r 2 )|2 v(r 1 − r 2 ) d3 r1 d3 r2 1 1 4π¯ h2 a 4π¯ h2 a |φ0 (r)|4 d3 r = m m (2π)3/2 a30 therefore ∆E = hω ¯

2 a . π a0

For a repulsive interaction (a > 0), there is an increase in the energy of the system. Conversely, in the case of an attractive interaction (a < 0), the ground state energy is lowered. (b) The perturbative approach yields a good approximation provided the ˆ Therefore, energy shift ∆E is small compared to the level spacing ¯hω of H. one must have |a| a0 , i.e. the scattering length must be small compared to the spreading of the ground state wave function. Section 19.3: Energy of a Bose–Einstein Condensate 19.3.1. Using the formulas provided in the text, one obtains: Ek (σ) = N

¯2 3 h 4 mσ 2

Ep (σ) = N

3 mω 2 σ 2 4

aa2 2 N (N − 1) ¯hω 30 . Eint (σ) = 2 π σ Indeed, there are N kinetic energy and potential energy terms, and N (N −1)/2 pairs which contribute to the interaction energy.

19.6 Solutions

199

19.3.2. With the change of variables introduced in the text, one ﬁnds

3 1 N −1 a 1 2 ˜ E(σ) = +σ ˜ + √ 2 4 σ ˜ ˜3 2π a0 σ so that

N −1 a . η= √ 2π a0 19.3.3. If the scattering length is zero, there is no interaction between the particles. The ground state of the system is the product of the N functions φ0 (r i ) and the ground state energy is E = (3/2)N ¯hω.

Section 19.4: Condensates with Repulsive Interactions ˜ σ ) as a function of σ 19.4.1. Figure 19.2 gives the variation of E(˜ ˜ for increasing values of η. The value of the function for a given value of σ ˜ increases as η ˜ does not depend on η. It is dominated increases. For large σ ˜ , the behavior of E 2 by the potential energy term 3˜ σ /4. ˜min increases as η increases. This minimum corresponds The minimum E the point where the potential energy term, which tends to favor small values of σ, matches the kinetic and interaction energy terms which, on the contrary, favor large sizes σ. Since the interactions are repulsive, the size of the system is larger than in the absence of interactions, and the corresponding energy is also increased.

˜ σ ) with σ Fig. 19.2. Variation of E(˜ ˜ for η = 0, 10, 100, 1000 (from bottom to top)

19.4.2. Let us assume η is much larger than 1 and let us neglect a priori σ 2 + η/˜ σ 3 is minimum for the kinetic energy term 1/˜ σ 2 . The function (3/4)˜ σ ˜min = (2η)1/5 where its value is ˜min = 5 (2η)2/5 . E 4 One can check a posteriori that it is legitimate to neglect the kinetic energy term 1/˜ σ 2 . In fact it is always smaller than one of the two other contributions ˜ to E:

200

19 Properties of a Bose–Einstein Condensate

• For σ ˜σ ˜min , one has 1/˜ 19.4.3. For a number of atoms N 1, the energy of the system as calculated by the variational method is 5 E = ¯ hω N 4

a 2 N π a0

2/5 .

(19.3)

This variation of E/N as N 2/5 is very well reproduced by the data. In Fig. 19.3 we have plotted a ﬁt of the data with this law. One ﬁnds E/N αN 2/5 with α = 8.2 × 10−33 Joule.

Fig. 19.3. Fit of the experimental data with an N 2/5 law

19.4.4. (a) One ﬁnds a0 = 1.76 µm and h ¯ ω = 9.4 10−32 Joule. (b) Consider the value a = 3.4 nm given in the text. The approximation η 1 will hold if N 1300. This is clearly the case for the data of Fig. 19.1. (c) The coeﬃcient α = 8.2 × 10−33 Joule found by ﬁtting the data leads to a = 2.8 nm. This value is somewhat lower than the expected value a = 3.4 nm. This is due to the fact that the result (19.3), E/(N ¯hω) 1.142 (N a/a0 )2/5 , obtained in a variational calculation using simple Gaussian trial functions, does not yield a suﬃciently accurate value of the ground state energy. With more appropriate trial wave functions, one can obtain, in the mean ﬁeld aphω) 1.055 (N a/a0 )2/5 . The ﬁt proximation and in the limit η 1: Egs /(N ¯ to the data is then in agreement with the experimental value of the scattering length.

19.6 Solutions

201

Section 19.5: Condensates with Attractive Interactions ˜ σ ) is represented in Fig. 19.4. We notice that it has 19.5.1. The function E(˜ a local minimum only for small enough values of η. For η < 0, there is always a minimum at 0, where the function tends to −∞. 10

E (σ ) 5 0 -5 -10

σ 0

1

2

3

˜ σ ) for η = 0; η = −0.1; η = −0.27; η = −1 (from top to Fig. 19.4. Plot of E(˜ bottom)

19.5.2. The absolute minimum at σ = 0 corresponds to a highly compressed atomic cloud. For such small sizes, approximation (19.2) for a “short range” potential loses its meaning. Physically, one must take into account the formation of molecules and/or atomic aggregates which have not been considered here. ˜ σ ) has an inﬂexion 19.5.3. The local minimum at σ ˜ = 0 disappears when E(˜ point where the derivative vanishes. This happens for a critical value of η determined by the two conditions: ˜ dE =0 d˜ σ

˜ d2 E =0. d˜ σ2

This leads to the system 2η 1 +1− 5 4 σ ˜ σ ˜ 3 8η 0 = 4 +1+ 5 σ ˜ σ ˜ 0=−

from which we obtain |ηc | =

2 55/4

0.27

σ ˜c =

1 51/4

0.67

202

19 Properties of a Bose–Einstein Condensate

or σc 0.67 a0 . If the local minimum exists, i.e. for |η| < |ηc |, one can hope to obtain a metastable condensate, whose size will be of the order of the minimum found in this variational approach. On the other hand, if one starts with a value of |η| which is too large, for instance by trying to gather too many atoms, the condensate will collapse, and molecules will form. 19.5.4. For the given experimental data one ﬁnds a0 = 3.1 µm, and a critical number of atoms: √ a0 ∼ 1400 , Nc = 2π ηc |a| in good agreement with experimental observations.

19.7 Comments The ﬁrst Bose–Einstein condensate of a dilute atomic gas was observed in Boulder (USA) in 1995 (M.H. Anderson, J.R. Ensher, M.R. Matthews, C.E. Wieman, and E.A. Cornell, Science 269, 198 (1995)) with rubidium atoms. The experimental data shown in this chapter for a sodium condensate come from the results published by M.-O. Mewes, M.R. Andrews, N.J. van Druten, D.M. Kurn, D.S. Durfee, and W. Ketterle, Phys. Rev. Lett. 77, 416 (1996). The measurement of the energy E/N is done by suddenly switching oﬀ the conﬁning potential and by measuring the resulting ballistic expansion. The motion of the atoms in this expansion essentially originates from the conversion of the potential energy of the atoms in the trap into kinetic energy. The experimental results on lithium have been reported by C. Bradley, C.A. Sackett, and R.G. Hulet, Phys. Rev. Lett. 78, 985 (1997). The Nobel prize 2001 has been awarded to E. Cornell, W. Ketterle, and C. Wieman for the achievement of Bose-Einstein condensation in dilute gases of alkali atoms, and for early fundamental studies of the properties of the condensates.

20 Magnetic Excitons

Quantum ﬁeld theory deals with systems possessing a large number of degrees of freedom. This chapter presents a simple model, where we study the magnetic excitations of a long chain of coupled spins. We show that one can associate the excited states of the system with quasi-particles that propagate along the chain. We recall that, for any integer k: N

e2iπkn/N = N if k = pN, with p integer;

n=1

= 0 otherwise.

20.1 The Molecule CsFeBr3 Consider a system with angular momentum equal to 1, i.e. j = 1 in the basis |j, m common to Jˆ2 and Jˆz . 20.1.1. What are the eigenvalues of Jˆ2 and Jˆz ? 20.1.2. For simplicity, we shall write |j, m = |σ, where σ = m = 1, 0, −1. Write the action of the operators Jˆ± = Jˆx ± iJˆy on the states |σ. 20.1.3. In the molecule Cs Fe Br3 , the ion Fe2+ has an intrinsic angular momentum, or spin, equal to 1. We write the corresponding observable Jˆ , and we note |σ the eigenstates of Jˆz . The molecule has a plane of symmetry, and the magnetic interaction Hamiltonian of the ion Fe2+ with the rest of the molecule is ˆ r = D Jˆ2 H D>0. z h2 ¯ ˆ r and the corresponding energy values? Are What are the eigenstates of H there degeneracies?

204

20 Magnetic Excitons

20.2 Spin–Spin Interactions in a Chain of Molecules We consider a one-dimensional closed chain made up with an even number N of Cs Fe Br3 molecules. We are only interested in the magnetic energy states of the chain, due to the magnetic interactions of the N Fe2+ ions, each with spin 1. We take {|σ1 , σ2 , · · · , σN }, σn = 1, 0, −1, to be the orthonormal basis of ˆn the states of the system; it is an eigenbasis of the operators {Jˆzn } where J is the spin operator of the n-th ion (n = 1, · · · , N ). ˆ = The magnetic Hamiltonian of the system is the sum of two terms H ˆ ˆ H0 + H1 where N ˆ0 = D H (Jˆzn )2 h2 n=1 ¯ ˆ 1 is a nearest-neighbor spin–spin interaction has been introduced in 1.3, and H term N ˆ n+1 ˆn · J ˆ1 = A J A>0. H h2 n=1 ¯ ˆ N +1 ≡ J ˆ 1. ˆ 1 , we deﬁne J To simplify the notation of H ˆ 0 (A D), ˆ 1 is a small perturbation compared to H We assume that H and we shall treat it in ﬁrst order perturbation theory. ˆ 0 , and give the cor20.2.1. Show that |σ1 , σ2 , · · · , σN is an eigenstate of H responding energy value. ˆ 0 ? Is it a degenerate level? 20.2.2. What is the ground state of H ˆ 0 ? What is the 20.2.3. What is the energy of the ﬁrst excited state of H 1 degeneracy d of this level? We shall denote by E the corresponding eigenspace ˆ 0 , of dimension d. of H ˆ 1 can be written as 20.2.4. Show that H N

1 ˆn ˆn+1 ˆ1 = A ˆn Jˆn+1 ) + Jˆn Jˆn+1 . J ( J H + J + − − + z z h2 n=1 2 ¯

20.3 Energy Levels of the Chain We now work in the subspace E 1 . We introduce the following notation |n, ± = |σ1 = 0, σ2 = 0, · · · , σn = ±1, σn+1 = 0, · · · , σN = 0. Owing to the periodicity of the chain, we deﬁne |N + 1, ± ≡ |1, ±.

20.3 Energy Levels of the Chain

205

20.3.1. Show that ˆ 1 |n, ± = A(|n − 1, ± + |n + 1, ±) + |ψn , H where |ψn is orthogonal to the subspace E 1 . Without giving the complete form of |ψn , give an example of one of its ˆ 0 to which |ψn belongs. components, and give the energy of the eigenspace of H 20.3.2. Consider the circular permutation operator Tˆ, and its adjoint Tˆ† , deﬁned by Tˆ|σ1 , σ2 , · · · , σN = |σN , σ1 , · · · , σN −1 Tˆ† |σ1 , σ2 , · · · , σN = |σ2 , σ3 , · · · , σN , σ1 . Write the action of Tˆ and Tˆ† on the states |n, ±. ˆ 1 and A(Tˆ + Tˆ† ) have the same 20.3.3. Check that, in the subspace E 1 , H matrix elements. 20.3.4. Show that the eigenvalues λk of Tˆ are the N -th roots of unity (we recall that N is assumed to be even): λk = e−iqk

qk = −π +

2kπ N

k = 0, · · · , N − 1 .

20.3.5. We seek, in E 1 , the 2N eigenvectors |qk , ± of Tˆ, each corresponding to an eigenvalue λk . Each |qk , ± is written cn (k) |n, ± . (20.1) |qk , ± = n

(a) Write a recursion relation between the coeﬃcients cn . (b) Show that 1 cn (k) = √ eiqk n N is a solution of this recursion relation.

(20.2)

(c) Show that the states |qk , ± deﬁned using (20.1) and (20.2) are orthonormal. (d) Show that the vectors |qk , ± are also eigenvectors of Tˆ† and Tˆ + Tˆ† , and give the corresponding eigenvalues. (e) Calculate the scalar product n, |qk , (, = ±) and write the expansion of the states |n, ± in the basis |qk , ±. ˆ 1 of Sect. 2 as a perturbation to H ˆ 0. 20.3.6. We treat the Hamiltonian H ˆ We limit ourselves to the ﬁrst excited level of H0 , and we want to calculate how the perturbation lifts the degeneracy of this level. We recall that, in the degenerate case, ﬁrst order perturbation theory consists in diagonalizing the restriction of the perturbing Hamiltonian in the degenerate subspace ˆ 0. of the dominant term H

206

20 Magnetic Excitons

(a) Explain why the results of questions 3.3 and 3.5 above allow one to solve this problem. (b) In ﬁrst order perturbation theory, give the new energy levels which arise ˆ 0 , and the corresponding eigenstates. from the ﬁrst excited state of H (c) Draw qualitatively the energies E(qk ) in terms of the variable qk which can be treated as a continuous variable, qk ∈ [−π, +π[, if N is very large. 1. What is the degeneracy of each new energy level?

20.4 Vibrations of the Chain: Excitons We now study the time evolution of the spin chain. 20.4.1. Suppose that at time t = 0, the system is in the state |Ψ (0) =

−1 N =± k=0

ϕk |qk ,

with

−1 N

|ϕk |2 = 1 .

=± k=0

Setting ω = 2A/¯ h, write the state |Ψ (t) at a later time t. 20.4.2. We assume that the initial state is |Ψ (0) = |qk , +. (a) Write the probability amplitude αn (t) and the probability Pn (t) of ﬁnding at time t the n-th spin pointing upwards, i.e. σn = +1 and σm = 0 for m = n. Show that Pn (t) is the same for all sites of the chain. (b) The molecules of the chain are located at xn = na, where a is the lattice spacing. Show that the probability amplitude αn (t) is equal to the value at x = xn of a monochromatic plane wave Ψk (x, t) = Cei(p(q)x−E(q)t)/¯h , where C is a constant, q = qk , and x is the abscissa along the chain. Express p(q) in terms of q. (c) Show that Ψk (x, t) is an eigenstate of the momentum operator pˆx = (¯ h/i)∂/∂x along the chain. Show that the value of p(q) ensures the periodicity of Ψk (x, t), i.e. Ψk (x + L, t) = Ψk (x, t), where L = N a is the length of the chain. (d) Show that, for |qk | 1, Ψk (x, t) satisﬁes a Schr¨odinger equation for a particle of negative mass m, placed in a constant potential; give the value of m. 20.4.3. In a more complete analysis, one can associate quasi-particles to the magnetic excitations of the chain. These quasi-particles, which we call “magnetic excitons”, have an energy E(qk ) and a momentum p(qk ). At very low temperatures, T ≈ 1.4 K, the chain is in the ground state ˆ 0 . If low energy neutrons collide with it, they can create excitons whose of H energy and momentum can be determined by measuring the recoil of the neutrons. The experimental result for E(q) as a function of q ∈ [−π, 0] is given in Fig. 20.1.

20.4 Vibrations of the Chain: Excitons

207

Fig. 20.1. Experimental measurement of the excitation energy E(q) as a function of q between −π and 0. The energy scale is in meV (10−3 eV)

(a) Deduce from that data approximate values for D and A. (b) What do you think of the approximation D A and of the comparison between theory and experiment? How could one improve the agreement between theory and experiment? (c) Is it justiﬁed to assume that the chain is in its ground state when it is at thermal equilibrium at 1.4 K? We recall the Boltzmann factor: N (E2 )/N (E1 ) = exp[−(E2 − E1 )/kT ], with k = 8.6 × 10−5 eV K−1 . 20.4.4. Consider, at time t = 0, the state |Ψ (0) =

N −1 k=0

ϕk |qk , +

with

N −1

|ϕk |2 = 1.

k=0

We assume that N 1, that the coeﬃcients ϕk have signiﬁcant values only in a close vicinity of some value k = k0 , or, equivalently, q ≈ q0 , and that, to a good approximation, in this vicinity, dE . E(q) = E(q0 ) + (q − q0 )u0 , u0 = dq q=q0 Show that the probability Pn (t) of ﬁnding σn = +1 at time t is the same as the probability Pn (t ) of ﬁnding σn = +1 at another time t whose value will be expressed in terms of t and of the distance between the sites n and n . Interpret the result as the propagation of a spin excitation wave along the chain. Calculate the propagation velocity of this wave and give its numerical value for a = 0.7 nm and q0 = −π/2. 20.4.5. We now assume that the initial state is |Ψ (0) = |n = 1, +.

208

20 Magnetic Excitons

(a) Write the probability Pm (t) of ﬁnding σm = +1 at a later time t? (b) Calculate the probabilities P1 (t) and P2 (t), in the case N = 2, and interpret the result. (c) Calculate P1 (t) in the case N = 8. Is the evolution of P1 (t) periodic? (d) For N 1, one can convert the above sums into integrals. The prob2 where the Jn (x) are the Bessel abilities are then Pm (t) ≈ |Jm−1 (ωt)| 2 (x)| = 1 and Jn = (−)n J−n . functions. These functions satisfy |J n

2 For x 1 we have Jn (x) ≈ πx cos(x − nπ/2 − π/4) if x > 2|n|/π, and Jn (x) ≈ 0 if x < 2|n|/π. Which sites are appreciably reached by the probability wave at a time t such that ωt 1? (e) Interpret the result as the propagation along the chain of a probability amplitude (or wave). Calculate the propagation velocity and compare it with the result obtained in question 4.4).

20.5 Solutions Section 20.1: The Molecule CsFeBr3 20.1.1. The results are: Jˆ2 : 2¯ h2 , Jˆz : m¯h ; m = 1, 0, −1. 20.1.2. One has: J+ |1 = 0√ J+ |0 =h ¯ √2|1 J+ | − 1 = h ¯ 2|0

√ =h ¯ √2|0 J− |1 J− |0 =h ¯ 2| − 1 J− | − 1 = 0 .

20.1.3. The eigenstates are the states |σ. The state |0 corresponds to the eigenvalue E = 0, whereas |+ and |−, which are degenerate, correspond to E = D. Section 20.2: Spin–Spin Interactions in a Chain of Molecules 20.2.1. It is straightforward to see that ˆ 0 |σ1 , σ2 · · · σN = D H

N n=1

the corresponding eigenvalue being E = D

(σn )2 |σ1 · · · σN ,

σn2 .

ˆ 0 corresponds to all the σn equal to zero, so 20.2.2. The ground state of H that E = 0. This ground state is non-degenerate. 20.2.3. The ﬁrst excited state corresponds to all the σ’s being zero except one: σn equal to ±1. The energy is D, and the degeneracy 2N , since there are N possible choices of the non-vanishing σn , and two values ±1 of σn . 20.2.4. J± = Jx ± iJy . A direct calculation leads to the result.

20.5 Solutions

209

Section 20.3: Energy Levels of the Chain 20.3.1. The action of the perturbing Hamiltonian on the basis states is, setting = ±: ˆ 1 |n, = A (|n − 1, + |n + 1, ) H +A (|0, · · · 0, σn = , 0 · · · 0, σn = −1, σn +1 = +1, 0 · · · 0 n =n

+|0 · · · 0, σn = , 0 · · · 0, σn = +1, σn +1 = −1, 0 · · · 0) . The vector |ψ = |σ1 = 1, σ2 = −1, 0 · · · 0, σn = , 0 · · · 0 belongs to this latter ˆ 0 with energy 3D. set; it is an eigenvector of H 20.3.2. The deﬁnition of Tˆ, Tˆ† and |n, ± implies: Tˆ|n, ± = |n + 1, ± ;

Tˆ† |n, ± = |n − 1, ± .

20.3.3. We therefore obtain A(Tˆ + Tˆ† )|n, ± = A(|n − 1, ± + |n + 1, ±) . Since ˆ 1 |n, ± = A(|n − 1, ± + |n + 1, ±) + |ψn H

where n , ±|ψn = 0 ,

ˆ 1 and A(Tˆ + Tˆ† ) obviously have the same matrix elements in the subspace H E 1. ˆ an eigenvalue λk satisﬁes λN = 1, which proves that 20.3.4. Since TˆN = I, k each eigenvalue is an N -th root of unity. Conversely, we will see in the following that each N -th root of unity is an eigenvalue. 20.3.5. (a) The corresponding eigenvectors satisfy |qk , ± =

cn |n, ±

Tˆ|qk , ± = λk |qk , ±

n

therefore one has

cn |n + 1, ± = λk

n

cn |n, ± .

n

Hence the recursion relation and its solution are λk cn = cn−1

cn =

1

c1 λn−1 k

= eiqk (n−1) c1 .

210

20 Magnetic Excitons

(b) The √ normalization condition n |cn |2 = 1 gives N |c1 |2 = 1. If we choose c1 = eiqk / N , the eigenvectors are of unit norm and we recover the solution given in the text of the problem. (c) The scalar product of |qk , and |qk , is easily calculated: qk , |qk , = δ,

1 2iπn(k−k )/N e = δ, δk,k . N n

(d) The vectors |qk , ± are eigenvectors of Tˆ† with the complex conjugate eigenvalues λ∗k . Therefore they are also eigenvectors of Tˆ + Tˆ† with the eigenvalue λk + λ∗k = 2 cos qk = −2 cos(2kπ/N ). (e) From the deﬁnition of the vectors, we have 1 n, |qk , = √ eiqk n δ N and (directly or by using the closure relation) N −1 1 −iqk n e |qk , ± . |n, ± = √ N k=0

ˆ 1 to the subspace E 1 is identical to A(Tˆ + Tˆ† ) 20.3.6. The restriction of H 1 (question 3.3). In E , the operator A(Tˆ + Tˆ† ) is diagonal in the basis |qk , ±. ˆ 1 is also diagonal in that basis. The energy levels Therefore the restriction of H are (20.3) E(qk ) = D + 2A cos(qk ) , corresponding to the states |qk , ±. As far as degeneracies are concerned, there is a twofold degeneracy for all levels (the spin value may be +1 or −1). In addition, for all levels except q = −π and q = 0, there is a degeneracy qk ↔ −qk (symmetry of the cosine). Therefore, in general, the degeneracy is 4. Section 20.4: Vibrations of the Chain: Excitons 20.4.1. At time t the state of the chain is (cf. (20.3)): |Ψ (t) = e−iDt/¯h ϕk e−iωt cos qk |qk , .

k

20.4.2. We now consider an initial state |qk , ±, evolving as e−iE(q)t/¯h |qk , ±. (a) We therefore obtain an amplitude 1 αn (t) = √ ei(qk n−E(qk )t/¯h) N and a probability Pn (t) = |αn |2 =

1 N,

which is the same on each site.

20.5 Solutions

211

(b) In the expression 1 αn (t) = √ ei(qk xn /a−E(qk )t/¯h) , N we see that αn (t) is the value at x = xn of the function Ψk (x, t) = √1 N

exp[i(px − Et)/¯ h] with E(q) = D + h ¯ ω cos q and p(q) = h ¯ q/a.

(c) The function Ψk (x) is an eigenstate of pˆx with the eigenvalue ¯hqk /a. Since N is even, we obtain: eiqk L/a = eiN qk = e2πik = 1 , which proves the periodicity of Ψk . (d) For |qk | 1, cos qk = 1 − qk2 /2 . Therefore E = E0 + p2 /2m with E0 = D + 2A

and

m=−

¯2 h ¯h =− 2 . 2Aa2 ωa

Ψk then satisﬁes the wave equation i¯ h

h2 ∂ 2 ψ ¯ ∂ψ =− + E0 ψ , ∂t 2m ∂x2

which is a Schr¨ odinger equation for a particle of negative mass (in solid state physics, this corresponds to the propagation of holes and in ﬁeld theory, to the propagation of anti-particles). 20.4.3. (a) With the data of the ﬁgure which resemble grosso modo the E(q) drawn in Fig. 20.2, one ﬁnds D + 2A ∼ 3.2 × 10−3 eV, and D − 2A ∼ 0.4 × 10−3 eV. Therefore: D ∼ 1.8 × 10−3 eV

A ∼ 0.7 × 10−3 eV .

(b) The approximation D A is poor. The theory is only meaningful to order (A/D)2 ∼ 10%. Second order perturbation theory is certainly necessary to account quantitatively for the experimental curve which has a steeper shape than a sinusoid in the vicinity of q = −π. (c) For T = 1.4 K, kT ∼ 1.2 × 10−4 eV, exp(−(D − 2A)/kT ) ∼ 0.04. To a few % , the system is in its ground state. 20.4.4. Approximating E(q) = E(q0 ) + (q − q0 )u0 in the vicinity of q0 , we obtain 1 ϕk ei(qk −q0 )(n−u0 t/¯h) . αn (t) = √ ei(q0 n−ω0 t) N k

212

20 Magnetic Excitons

Fig. 20.2. Energy levels

Since the global phase factor does not contribute to the probability, one has Pn (t) = Pn (t ) with h ¯ t = t + (n − n) . u0 This corresponds to the propagation of a wave along the chain, with a group velocity a dE u0 a 2a A = sin q0 . =− vg = h ¯ h dq q=q0 ¯ ¯h For q0 = −π/2 and a = 0.7 nm, we ﬁnd vg ∼ 1500 ms−1 . One can also evaluate u0 ∼ 1.2 meV directly on the experimental curve, which leads to vg ∼ 1300 ms−1 . √ −iqk 20.4.5. If |Ψ (0) = |n = 1, +, then ϕ+ / N and ϕ− k =e k = 0. (a) The probability is Pm (t) = |m, +|Ψ (t)|2 , where m, +|Ψ (t) =

e−iDt/¯h iqk (m−1) −iωt cos qk e e . N k

(b) N=2: There are two possible values for qk : q0 = −π and q1 = 0. This leads to P1 (t) = cos2 ωt, P2 (t) = sin2 ωt. These are the usual oscillations of a twostate system, such as the inversion of the ammonia molecule. (c) N=8: qk cos(qk )

−π

− 3π 4

−1

− √12

− π2

− π4

0

0

√1 2

1

π 4 √1 2

π 2

0

3π 4 − √12

The probability P1 of ﬁnding the excitation on the initial site is P1 (t) =

√ 2 1 2 cos (ωt/2) + cos(ωt/ 2) . 4

The system is no longer periodic in time. There cannot exist t = 0 for which

20.5 Solutions

213

√ P1 (t) = 1, otherwise there would exist n and n such that 2 = n /n. (d) Since Jn (ωt) ∼ 0 for ωt < 2|n|/π, only sites for which |m − 1| < πωt/2 are reached at time t. For large ωt, the amplitude is the same for all sites of the same parity: Pm (t) =

2 π π cos2 ωt − (m − 1) − . πωt 2 4

We notice in particular that Pm (t) + Pm+1 (t) = 2/(πωt) is independent of m and varies slowly with t. (e) The probability wave becomes delocalized very quickly on the chain (ωt > a few π). The edges of the region where the probability is non zero propagate in opposite directions with the velocity v = πωa/2. This is comparable with what we have found in 4.4 for a wave packet near q = π/2. The experimental data displayed in this chapter were obtained by B. Dorner et al., Z. Phys. B 72, 487 (1988).

21 A Quantum Box

In recent years, it has become possible to devise quantum boxes (also called quantum dots) of nanometric dimensions, inside which the conduction electrons of a solid are conﬁned at low temperatures. The ensuing possibility to control the energy levels of such devices leads to very interesting applications in micro-electronics and opto-electronics. A quantum box is made of a material A on which another material B is deposited. A set of quantum boxes is shown on Fig. 21.1. The dots of In As (material B) are deposited on a substate of Ga As (material A).

Fig. 21.1. Picture of a set of quantum boxes obtained with a tunneling microscope. The side of the square is 1 µm long and the vertical scale will be studied below

216

21 A Quantum Box

In this chapter, we are interested in the motion of an electron in a twodimensional box. We note −q the electric charge of the electron, and we neglect spin eﬀects. We shall assume that in a solid, the dynamics of an electron is described by the usual Schr¨ odinger equation where: (i) the mass of the electron is replaced by an eﬀective mass µ, (ii) the atoms of the materials A and B create an eﬀective potential V (x, y) which is slowly varying on the atomic scale.

21.1 Results on the One-Dimensional Harmonic Oscillator Consider a particle of mass µ placed in the one-dimensional potential V (x) = µω 2 x2 /2. We recall the deﬁnition of the annihilation and creation operators ˆ†x of the oscillator in terms of the position and momentum operators a ˆx and a x ˆ and pˆx

pˆx pˆx µω µω 1 1 † + i√ − i√ a ˆx = √ a ˆx = √ . (21.1) x ˆ x ˆ h ¯ ¯h hµω ¯ ¯hµω 2 2 The Hamiltonian of the system can be written as

2 1 ˆ x = pˆx + 1 µω 2 x H ˆ2 = h ¯ω n ˆx + where 2µ 2 2

ˆ†x a ˆx . n ˆx = a

(21.2)

We also recall that the eigenvalues of the number operator n ˆ x are the nonnegative integers. Noting |nx the eigenvector corresponding to the eigenvalue nx , we have √ √ a ˆx |nx = nx |nx − 1 (21.3) a ˆ†x |nx = nx + 1|nx + 1 21.1.1. We recall that the ground state wave function is

µω 1/4 µωx2 ψ0 (x) = exp − . π¯ h 2¯h What is the characteristic extension 0 of the electron’s position distribution in this state? 21.1.2. The eﬀective mass µ of the electron in the quantum box is µ = 0.07 m0 , where m0 is the electron mass in vacuum. We assume that ¯hω = 0.060 eV, i.e. ω/(2π) = 1.45 × 1013 Hz. (a) Evaluate 0 numerically. (b) At a temperature of 10 Kelvin, how many levels of the oscillator are populated signiﬁcantly? (c) What is the absorption wavelength of radiation in a transition between two consecutive levels?

21.2 The Quantum Box

217

21.2 The Quantum Box We assume that the eﬀective two-dimensional potential seen by an electron in the quantum box is: V (x, y) =

1 2 2 µω x + y 2 . 2

(21.4)

ˆ 0 = (ˆ We note H p2x + pˆ2y )/2µ + V (x, y) the Hamiltonian of the electron. ˆ†y and n ˆ y in an analogous way to (21.1) 21.2.1. We deﬁne the operators a ˆy , a and (21.2). Give a justiﬁcation for the fact that the states |nx , ny , which ˆ y with integer eigenvalues nx and ny , form an are eigenstates of n ˆ x and n ˆ 0 in terms of nx and ny . ˆ 0 . Give the energy levels EN of H eigenbasis of H 21.2.2. What is the degeneracy gN of each level EN where N = 0, 1, 2 . . . ? ˆz = x 21.2.3. Express the operator L ˆpˆy − yˆpˆx as a function of the operators † † ˆx , a ˆy , a ˆy . a ˆx , a ˆ 0 . Do the states ˆ z on the eigenstates |nx , ny of H 21.2.4. Write the action of L |nx , ny have a well-deﬁned angular momentum Lz ? ˆ 0. 21.2.5. We are now interested in ﬁnding another eigenbasis of H ˆ ˆ (a) Show that H0 and Lz commute. Interpret this result physically. (b) We introduce the “left” and “right” annihilation operators as 1 ax + iˆ a ˆl = √ (ˆ ay ) 2

1 ax − iˆ a ˆr = √ (ˆ ay ) 2

(21.5)

ˆ†r . Write the commutation and the corresponding creation operators a ˆ†l , a relations of these four operators. ˆ†l a ˆl and n ˆr = a ˆ†r a ˆr commute. Using the values of the (c) Show that n ˆl = a † commutators [ˆ al , a ˆl ] and [ˆ ar , a ˆ†r ], and following the same procedure as in the usual quantization of the harmonic oscillator, justify that the eigenˆ l and n ˆ r are integers. values nl and nr of n ˆ r } form a complete set of commuting observables in We assume that {ˆ nl , n the problem under consideration, and we note {|nl , nr } the corresponding eigenbasis. ˆ r in terms of a ˆx , a ˆ†x , a ˆy and a ˆ†y . Deduce (d) Write the expression of n ˆ l and n ˆ z in terms of n ˆ 0 and L from that an expression of the operators H ˆ l and n ˆr . ˆ 0 and Justify that the states {|nl , nr } form an eigenbasis common to H ˆz. L ˆ z and H ˆ 0 . What are the allowed (e) We note m¯h and EN the eigenvalues of L values for the quantum number m in a given energy level EN ? (f) Represent the allowed couples of quantum numbers (m, N ) by points in the (Lz , EN ) plane. Show that one recovers the degeneracy of the levels ˆ 0. of H

218

21 A Quantum Box

ˆ 0 generated by |nx = 1, ny = 0 21.2.6. We consider the eigen-subspace of H and |nx = 0, ny = 1 as deﬁned in Sect. 2.1 above. Write the eigenstates of ˆ z in this basis and give the corresponding eigenvalues. L

21.3 Quantum Box in a Magnetic Field One applies a uniform magnetic ﬁeld B, parallel to the z axis, on the quantum box. This ﬁeld derives from the vector potential A(r) = −(yB/2)ux + (xB/2)uy where ux and uy are the unit vectors along the x and y directions. We assume that the Hamiltonian of the electron in the quantum box and in presence of the ﬁeld is given by 1 1 ˆ B = 1 (ˆ H px + qAx (ˆ (ˆ py + qAy (ˆ r ))2 + r ))2 + µω 2 (ˆ x2 + yˆ2 ) 2µ 2µ 2 with the usual canonical commutation relations [ˆ x, pˆx ] = [ˆ y , pˆy ] = i¯h and y , pˆx ] = 0. We introduce the cyclotron angular frequency ωc = qB/µ, [ˆ x, pˆy ] = [ˆ (ωc > 0). ˆ B and show that one can always ﬁnd a basis of the sys21.3.1. Expand H tem for which both the total energy and the angular momentum along z are simultaneously well deﬁned. 21.3.2. We deﬁne Ω = ω 2 + ωc2 /4. By redeﬁning in a simple manner the operators of Sect. 2, give the energy levels Enl ,nr of the system in terms of the two integers nl and nr . 21.3.3. For simplicity we subtract the zero point energy ¯hΩ from the energies ˜n ,n = En ,n − ¯ hΩ. Enl ,nr , and we set E r r l l ˜n ,n in the two limits of (a) Give the approximate expressions of the levels E r l a weak and of a strong ﬁeld, and give a deﬁnition of these regimes. (b) Plot as a function of the magnetic ﬁeld, the positions of the energy levels originating from the N = 0, 1, 2 levels in the absence of magnetic √ ﬁeld. (c) Show that two levels cross for a value of B such that ωc = ω/ 2 and specify which states they correspond to. √ 21.3.4. In the following we assume that ωc < ω/ 2. Using the values of ω and of the eﬀective mass µ given in the ﬁrst section of the problem, determine which values of the magnetic ﬁeld correspond to this inequality. ˆ B have respective energies 21.3.5. Show that the ﬁrst three eigenstates of H ¯Ω , E0 = h

E− = 2¯ hΩ −

hωc ¯ , 2

E+ = 2¯ hΩ +

¯hωc . 2

We note |u0 , |u− , |u+ the three corresponding eigenstates. What is the value of the angular momentum Lz in each of these three states?

21.4 Experimental Veriﬁcation

219

21.4 Experimental Veriﬁcation One can study the energy levels of an electron in a quantum box by measuring the absorption spectrum of a light beam. Absorption peaks appear at the Bohr frequencies (Ef −Ei )/h of the box, corresponding to excitations of the electron from an initial level |ui to a ﬁnal one |uf . 21.4.1. At a temperature of 10 K, one observes that only the level |ui = |u0 contributes signiﬁcantly to the absorption signal. Justify this fact using the result of question 1.2. 21.4.2. The experimental values of the frequencies of the ﬁrst two absorption peaks of a quantum box are given in Fig. 21.2, according to the values of the applied magnetic ﬁeld. Verify that the model developed above accounts for the slope of the curves for suﬃciently large values of B, but that it fails to describe the behavior near B = 0.

absorption frequency (1012 Hz)

17 16 15 14 13 12 0

2

4

6

8

10

Magnetic field (Tesla) Fig. 21.2. Frequencies ν = ω/2π of the two ﬁrst absorption peaks of a quantum box as a function of the magnetic ﬁeld B

21.4.3. Role of the z Dimension The conﬁnement along the z direction can be simulated by an inﬁnite onedimensional square well potential of extension D in this direction. (a) Recall the energy levels of a particle of mass µ in a one-dimensional inﬁnite well of width D. What is the energy diﬀerence between the two lowest lying states?

220

21 A Quantum Box

(b) Under what condition between D and ω is it legitimate to consider that the motion along the z axis is “frozen”, and to restrict to the lowest lying levels of the harmonic motion in the x, y plane? (c) On the picture 21.1, the vertical (z) and horizontal (xy) scales are diﬀerent. In order to neglect the motion along the z direction, which of the two scales must be contracted? One can make use of the value of l0 calculated in Sect. 1.2.

21.5 Anisotropy of a Quantum Box In order to reproduce the positions of the absorption peaks for low values of the ﬁeld (see Fig. 21.2), we assume that the conﬁning potential in the quantum box is slightly anisotropic. We therefore replace the expression (21.4) of the conﬁning potential by V (x, y) =

1 1 2 µω (1 + )x2 + µω 2 (1 − )y 2 2 2

with 1 .

(21.6)

We treat the problem using perturbation theory. In what follows, we asˆ B, can sume that the magnetic ﬁeld is small (ωc ω). The Hamiltonian H therefore be written as: ˆ B, = H ˆ0 + W ˆ H with 2 ˆ2 1 ˆ z + µω (ˆ ˆ0 = p ˆ = ωc L H + µω 2 (ˆ x2 − yˆ2 ) . x2 + yˆ2 ) and W 2µ 2 2 2

ˆ is the perturbation, and we neglect terms of order B 2 . In this expression, W We work in the basis {|nx , ny }. 21.5.1. Using the potential given in (21.6), determine the energy levels of the electron for B = 0. 21.5.2. Applying perturbation theory, evaluate to ﬁrst order in B and the energy shift of the ground state of the quantum box with respect to the value found in Sect. 2. 21.5.3. We are now interested in the position of the two energy states origiˆ 0 found in Sect. 2. nating from the ﬁrst excited level of H (a) Write the restriction of the Hamiltonian HB, in the basis {|nx = 1, ny = 0 , |nx = 0, ny = 1} of the corresponding subspace. (b) Deduce the approximate values of the energies of interest E− (B, ) and E+ (B, ) (with E− < E+ ). (c) Give the expressions of the corresponding eigenstates |u− (B, ) and |u+ (B, ). One can introduce the mixing angle α such that tan(2α) = ωc /(ω).

21.6 Solutions

221

21.5.4. We now turn back to the experimental data of Fig. 21.2. (a) Is the transition region between large values of B and B = 0 correctly described by this model of an anisotropic quantum box? (b) What value of the anisotropy can be deduced from the positions of the ﬁrst two absorption peaks in the absence of a magnetic ﬁeld?

21.6 Solutions Section 21.1: Results on the One-Dimensional Harmonic Oscillator 21.1.1. The characteristic spatial√extension of the position distribution is h/(µω). More precisely, 0 / 2 is the r.m.s. deviation of the probability 0 = ¯ law |ψ0 (x)|2 for the position distribution. 21.1.2. (a) One ﬁnds 0 = 4.3 nm. (b) The ratio of populations in the ﬁrst excited level n = 1 and the ground state n = 0 is given by the Boltzmann law r = exp(−¯ hω/(kB T )) = exp(−70) = 5 × 10−31 , which is negligible. The population of other excited levels is even smaller. Therefore, at T = 10 K, only the ground state is populated. (c) One ﬁnds λ = 2πc/ω = 21 µm, which corresponds to an infrared radiation. Section 21.2: The Quantum Box ˆ0 = h 21.2.1. The Hamiltonian is H ¯ ω(ˆ nx + n ˆ y + 1), and the operators n ˆ x and n ˆ y commute. We can ﬁnd an eigenbasis common to the two operators. If a x dependence is completely defunction Ψ (x, y) is an eigenfunction of n ˆ x , its termined (Hermite function of the variable x µω/¯h, corresponding to |nx ). ˆ y } is therefore a CSCO, with Similarly, for the y dependence. The set {ˆ nx , n ˆ 0 , the eigenvalue coreigenbasis |nx , ny . This basis is also an eigenbasis of H ¯ ω(N + 1) with N = nx + ny . responding to |nx , ny is EN = h 21.2.2. The energy level EN = h ¯ ω(N + 1) corresponds to N + 1 possible couples for (nx , ny ): (N, 0), (N − 1, 1), . . ., (0, N ). The degeneracy is therefore gN = N + 1.

222

21 A Quantum Box

21.2.3. We replace x ˆ, pˆx , yˆ, pˆy by their values:

h † ¯ hµω † ¯ a ˆx + a a ˆx − a pˆx = i ˆx ˆx x ˆ= 2µω 2

...

which leads to:

†

†

†

h † ˆ z = i¯ L a ˆx + a ˆy − a ˆy + a ˆx − a ˆx a ˆx a ˆy − a ˆy a ˆx = i¯h a ˆ†y − a ˆ†x a ˆy . 2 21.2.4. Making use of the action of the creation and annihilation operators, one ﬁnds: ˆ z |nx , ny = i¯ L h nx (ny + 1) |nx − 1, ny + 1 − (nx + 1)ny |nx + 1, ny − 1 . ˆ z and does not have a A state |nx , ny is not, in general, an eigenstate of L well deﬁned angular momentum. The only exception is the state for which ˆ z |0, 0 = 0. It is a state with Lz = 0. nx = ny = 0, and L ˆ 0. 21.2.5. We are now interested in ﬁnding another eigenbasis of H ˆ ˆ (a) The commutator of H0 and Lz can be calculated using their expressions in terms of the creation and annihilation operators. We ﬁrst calculate ˆ†x ] = a ˆ†x and [ˆ nx , a ˆx ] = −ˆ ax , and the analogous relations for y. One [ˆ nx , a obtains: ˆ 0, L ˆ z ] = i¯ [H h2 ω [ˆ nx + n ˆy , a ˆx a ˆ†y − a ˆ†x a ˆy ]

nx , a ˆx ]a†y − [ˆ nx , a ˆ†x ]ˆ ay + a ˆx [ˆ ny , a†y ] − a ˆ†x [ˆ ny , a ˆy ] = i¯ h2 ω [ˆ

ax a†y − a ˆ†x a ˆy + a ˆx a†y + a ˆ†x a ˆy = i¯ h2 ω −ˆ =0. This result is the consequence of the rotation invariance around the z axis of the potential V (x, y). This can also be proven by using the polar coordinate ∂ ˆ z = −i¯ ˆ0 expression: L h ∂φ , which commutes with the two contributions to H (kinetic and potential energy). (b) There are 4 operators, and therefore 6 commutators to evaluate. We ﬁrst remark that the two creation operators commute, and so do the two annihilation operators. Therefore: ˆ†r ] = 0 , [ˆ a†l , a A simple calculation leads to: 1 a ˆl , a ax + iˆ ˆ†l = [ˆ ay , a ˆ†x − iˆ a†y ] = 2 # 1 " ax − iˆ ˆ†r = [ˆ ay , a ˆ†x + iˆ a†y ] = a ˆr , a 2

[ˆ al , a ˆr ] = 0 .

1 [ˆ ax , a ˆ†x ] + 2 1 [ˆ ax , a ˆ†x ] + 2

1 [ˆ ay , a ˆ†y ] = 1 , 2 1 [ˆ ay , a ˆ†y ] = 1 . 2

21.6 Solutions

223

Finally, one ﬁnds: # 1 " ax + iˆ ˆ†r = [ˆ ay , a ˆ†x + iˆ a†y ] = a ˆl , a 2 1 a ˆr , a ax − iˆ ˆ†l = [ˆ ay , a ˆ†x − iˆ a†y ] = 2

1 [ˆ ax , a ˆ†x ] − 2 1 [ˆ ax , a ˆ†x ] − 2

1 [ˆ ay , a ˆ†y ] = 0 , 2 1 [ˆ ay , a ˆ†y ] = 0 . 2

Therefore, any “right” and “left” operators commute. The commutation relations between right (or left) creation and annihilation operators are the same as for usual one-dimensional creation and annihilation operators. ˆ r is obvious, (c) The commutation between the hermitian operators n ˆ l and n since any left operator commutes with any right operator. The quantization of the one-dimensional harmonic oscillator is entirely based on the commutation ˆ† a ˆ are relation [ˆ a, a ˆ† ] = 1, which leads to the fact that the eigenvalues of a ˆr , the non-negative integers. The same holds here for the operators n ˆ l and n the corresponding eigenvalues are the couples of integers (nl , nr ). As it is suggested in the text, we assume for the moment that the common eigenbasis ˆ r is unique, and we note |nl , nr the eigenvector corresponding to of n ˆ l and n the couple of eigenvalues (nl , nr ). (d) One ﬁnds:

1 † 1 (ˆ ax − iˆ n ˆx + n a†y ) (ˆ ax + iˆ ay ) = ˆ y + i(ˆ a†x a ˆy − a ˆx a ˆ†y ) 2 2 ˆ ˆ Lz 1 H0 −1− , = 2 ¯ hω h ¯

1 † 1 ax + iˆ n ˆx + n n ˆ r = (ˆ a†y ) (ˆ ax − iˆ ay ) = ˆ y − i(ˆ a†x a ˆy − a ˆx a ˆ†y ) 2 2 ˆ0 ˆz L 1 H −1+ , = 2 ¯ hω h ¯ n ˆl =

therefore ˆ0 = h ˆz = h H ¯ ω (ˆ nl + n ˆ r + 1) , L ¯ (ˆ nr − n ˆl) . ˆ 0 and L ˆ z can be expressed in terms of the sole operators The operators H ˆ r . The eigenbasis common to n ˆ l and n ˆ r is therefore also a common n ˆ l and n ˆz. ˆ 0 and L eigenbasis of H ˆ 0 and L ˆ z , with eigenvalues E = (e) The vector |nl , nr is the eigenstate of H ¯ (nr − nl ). We therefore have N = nl + nr and hω(nl + nr + 1) and Lz = h ¯ m = nr − nl . We recover the integer values for the orbital angular momentum, as expected. For a given N , the value of m belongs to the set {−N, −N + 2, . . . , N − 2, N }, therefore there are N + 1 possible values. We remark that in an energy level EN , m has the same parity as N . This comes from the parity invariance of the problem under consideration. (f ) The graphical representation is given on Fig. 21.3. On a given horizontal line corresponding to a given energy value, one ﬁnds N + 1 points, which

224

21 A Quantum Box

E / hω 4 3 2 1 Lz / h -3

-2

-1

0

1

2

3

Fig. 21.3. Allowed quantum numbers for the couple Lz , E

ˆ 0 found previously. This correspond to the degeneracy of an energy level of H ˆ r } form a CSCO. If two diﬀerent states would justiﬁes the fact that {ˆ nl , n correspond to the same couple of eigenvalues (nl , nr ), the corresponding point of the diagram would be twofold degenerate, and the degeneracy of the energy level EN would be larger than N + 1. ˆ z corresponding to 21.2.6. We must ﬁnd in this subspace two eigenvectors of L the two eigenvalues ±¯ h. A ﬁrst method for ﬁnding these eigenvectors consists ˆ z on the vectors of the basis {|nx , ny }. Using in calculating the action of L ˆ z in terms of a ˆx , a ˆy , . . ., one ﬁnds: the expression of L ˆ z |nx = 1, ny = 0 = i¯ L h |nx = 0, ny = 1 ˆ z |nx = 0, ny = 1 = −i¯ L h |nx = 1, ny = 0 ,

0 −i¯ h or the 2 × 2 matrix to diagonalize: . i¯ h 0 The eigenstates associated to the eigenvalues ±¯ h are therefore: √ (|nx = 1, ny = 0 ± i|nx = 0, ny = 1) / 2 . Another method consists in starting from the ground state |nl = 0, nr = 0 and letting act on this state: (i) the operator a ˆ†r in order to obtain the eigenvector of energy 2¯hω and angular momentum +¯ h, (ii) the operator a ˆ†l in order to obtain the eigenvector of energy 2¯ hω and angular momentum −¯ h. Of course, we recover the previous result.

21.6 Solutions

225

Section 21.3: Quantum Box in a Magnetic Field ˆ B , one ﬁnds: 21.3.1. By expanding H ˆB = H

pˆ2x + pˆ2y + 2µ

µω 2 q2 B 2 + 2 8

(ˆ x2 + yˆ2 ) +

ˆz ωc L . 2

ˆ z /2, ˆB = H ˆ (Ω) + ωc L 21.3.2. If we set Ω = ω 2 + ωc2 /4, we can rewrite H 0 (Ω) ˆ where H0 is the Hamiltonian of a two-dimensional oscillator of frequency Ω: pˆ2 + pˆ2y µΩ 2 2 ˆ (Ω) = x + (ˆ x + yˆ2 ) . H 0 2µ 2 One can then repeat the method of the previous section, by replacing ω by Ω in ˆy ,... One constructs an eigenbasis common the deﬁnition of the operators a ˆx , a (Ω) ˆ z , which we continue to note {|nl , nr }, the eigenvalues being to H0 and L ˆB, hΩ(nl + nr + 1) and m¯h. Each vector |nl , nr is also an eigenvector of H ¯ corresponding to the energy Enl ,nr = h ¯ Ω(nl + nr + 1) + h ¯ ωc (nr − nl )/2 ωc ωc ¯ Ω− ¯Ω . =h ¯ Ω+ nr + h nl + h 2 2 21.3.3. (a) Two limiting regimes of the magnetic ﬁeld can be considered, corresponding to the limits ωc ω (very weak magnetic ﬁeld) and ωc ω (very strong magnetic ﬁeld). In the ﬁrst case, we have in ﬁrst order in B: ˜n ,n ¯ E hω(nl + nr ) + h ¯ ωc (nr − nl )/2 , r l which corresponds to a linear variation in B of the N + 1 levels arising from the level EN in the absence of the ﬁeld. The slope (¯hqB/(2µ) (nr − nl ) is diﬀerent for each level, which means that there is no degeneracy if B does not vanish. For strong ﬁelds, one ﬁnds Ω+

ωc ωc , 2

Ω−

ω2 ωc ω . 2 ωc

We therefore have ˜n ,n ¯ E hωc nr r l

if nr = 0 ,

and

¯ ω2 ˜n ,0 h E nl . l ωc

For a non-vanishing nr , the energy level increases linearly with B, the slope ˜ tends to zero as 1/B. being proportional to nr . For nr = 0, the energy E ˜ (b) The energy levels Enl ,nr corresponding to N = 0, 1, 2 are represented on Fig. 21.4.

226

21 A Quantum Box

E/hΩ 2

1

0

ω /ω c 0

1

2

3

4

5

˜n ,nr arising from N = 0, 1, 2 as a function Fig. 21.4. Variation of the energy levels E l of the magnetic ﬁeld B

(c) We notice on Fig. 21.4 that the levels nl = 2, nr = 0 and nl = 0, nr = 1 cross each other. The corresponding value of the ﬁeld B is given by the solution of the equation ωc ωc =2 Ω− Ω+ , 2 2 √ or 3 ωc = 2 Ω, i.e. ωc = ω/ 2. √ 21.3.4. √ The value of the ﬁeld B which corresponds to ωc = ω/ 2 is µω/(q 2) 26 T. 21.3.5. If we assume that the ﬁeld B is smaller than 21 T, the three ﬁrst ˆ B correspond to nl = nr = 0 (ground state of energy energy levels of H hΩ − ¯ hωc /2), and nl = 0, nr = 1 (energy hΩ), nl = 1, nr = 0 (energy 2¯ ¯ ˆ z with the eigenvalues 2¯hΩ + h ¯ ωc /2). These three states are eigenstates of L 0, −¯ h and h ¯ respectively. Section 21.4: Experimental Veriﬁcation 21.4.1. In the absence of a magnetic ﬁeld, we saw in question 1.2 that only the level nx = ny = 0 is appreciably populated for T = 10 K. As the magnetic ﬁeld increases, the energy splitting between the ground state and the ﬁrst excited √ state diminishes√but it stays much smaller than kB T if ωc is less √ ¯ ω/ 2. For that value, the ratio than ω/ 2. For ωc = ω/ 2, the splitting is h between the populations of the ﬁrst excited state and the ground state is r = exp(−49) = 3.7 × 10−22 . Since only the ground state is populated, all the detectable absorption lines will occur from transition starting from this state. 21.4.2. The ﬁrst two absorption peaks correspond to the transitions |u0 ↔ |u− and |u0 ↔ |u+ . The corresponding frequencies ν± are such that ν± =

ωc Ω ± . 2π 4π

21.6 Solutions

227

The domain in B explored on the experimental ﬁgure of the text corresponds to values of ωc which are small compared to ω. We can therefore use the weak-ﬁeld expansion of question 3.3 in order to simplify this expression ωc ω ± . ν± = 2π 4π We therefore expect that the frequencies ν± will vary linearly with B, the slopes being ±q/(4πµ), and that the two straight lines will cross for a vanishing ﬁeld at the frequency ω/(2π). This linear variation of ν± does appear on the ﬁgure for higher values of B and the measured slope is close to the expected value (2 × 1011 Hz T−1 ). However, the experimentally observed behavior for a very weak ﬁeld does not correspond to our theoretical prediction. Instead of two lines of same frequency for B = 0, there is a ﬁnite diﬀerence ν+ − ν− . 21.4.3. Role of the z Dimension (a) The energy levels of an inﬁnite square well of size D are given by En = π 2 ¯h2 n2 /(2µD2 ), with n positive integer, the corresponding eigenstates are the functions χn (z) ∝ sin(nπz/D). The splitting between the ground state and h2 /(2µD2 ). the ﬁrst excited state is ∆E = 3π 2 ¯ (b) In order to consider that the motion along z is “frozen”, the energy splitting ∆E between the two ﬁrst levels of the square well must be very large compared to ¯hω. If this condition is satisﬁed, the accessible states for the electron conﬁned in the quantum box (in a reasonable domain of temperatures and exciting frequencies) will be simply combinations of the vectors |nx , ny ⊗ |χ0 . It is then legitimate to neglect the dynamics of the electron along z. If this condition is not satisﬁed, absorption lines can appear for frequencies near those presented on the experimental ﬁgure. They will correspond to the excitation of the motion along the z axis. The condition that the z motion be “frozen” is h2 3π 2 ¯ ¯ hω , or equivalently D π0 . (21.7) 2µD2 (c) In order for the harmonic approximation of the transverse motion to be valid, the transverse extension ∆L of the quantum box must be large compared to 0 . The condition obtained in the previous question D π0 , put together with 0 ∆L, imposes that the box must have a very ﬂat geometrical shape: the height D along z must be very small compared to its transverse extension in xy. We conclude that the vertical scale of the picture 21.1 is very dilated. Section 21.5: Anisotropy of a Quantum Box ˆ y with ˆx + H 21.5.1. For vanishing B, the Hamiltonian is H 2 ˆ x = pˆx + 1 µω 2 (1 + )ˆ H x2 , 2µ 2

ˆy = H

pˆ2y 1 + µω 2 (1 − )ˆ x2 . 2µ 2

228

21 A Quantum Box

ˆ x and H ˆ y , corresponding to products One can ﬁnd a common eigenbasis for H of Hermite functions in the variable x µω(1 + )/¯ h by Hermite functions in the variable y µω(1 − )/¯ h. The corresponding eigenvalues are √ √ ¯ hω (nx −ny ) hω 1 − (ny +1/2) ¯hω (nx +ny +1)+ hω 1 + (nx +1/2)+¯ ¯ 2 where nx , ny are non-negative integers. 21.5.2. To ﬁrst order in B and , the shift of the ground state energy is given by the matrix element 2 ˆ z |0, 0 + µω 0, 0|ˆ ˆ |0, 0 = ωc 0, 0|L x2 − yˆ2 |0, 0 . ∆E0,0 = 0, 0|W 2 2

ˆ z with eigenvalue 0. The ﬁrst term in this The state |0, 0 is an eigenstate of L y 2 |0, 0, sum therefore vanishes. By symmetry, we have 0, 0|ˆ x2 |0, 0 = 0, 0|ˆ which means that the second term also vanishes, to ﬁrst order in and B. ˆ z in the basis under 21.5.3. (a) We have already determined the matrix L consideration in question 2.6. We must calculate the matrix elements of x ˆ2 2 and yˆ . In order to do that, the simplest is to use the expressions of x ˆ and yˆ in terms of creation and annihilation operators. One has: x ˆ2 =

h ¯ (ˆ ax + a ˆ†x )(ˆ ax + a ˆ†x ) , 2µω

which leads to h ¯ 1, 0|ˆ a†x a ˆx + a ˆx a ˆ†x |1, 0 2µω 3¯h h ¯ (1 + 2) = , = 2µω 2µω 1, 0|ˆ x2 |0, 1 = 0, 1|ˆ x2 |1, 0 = 0 , h ¯ 0, 1|ˆ x2 |0, 1 = 0, 1|ˆ ax a ˆ†x |0, 1 2µω h ¯ , = 2µω

1, 0|ˆ x2 |1, 0 =

where we have set |0, 1 ≡ |nx = 0, ny = 1, etc. for simplicity. We obtain a ˆ B, in similar result by exchanging the roles of x and y. The restriction of H the subspace of interest is therefore

h ω −iωc ¯ ˆ hω + . [HB, ] = 2¯ 2 iωc − ω (b) The energy eigenvalues are obtained by diagonalizing this 2 × 2 matrix hω ± E± (B, ) = 2¯

¯ 2 2 h ω + ωc2 . 2

21.7 Comments

229

(c) Setting tan 2α = ωc /( ω), the above matrix is written as:

h 2 2 ¯ cos 2α −i sin 2α 2 ˆ hω + ω + ωc , [HB, ] = 2¯ i sin 2α − cos 2α 2 whose eigenvectors are

|u− (B, ) =

i sin α cos α

|u+ (B, ) =

cos α i sin α

.

21.5.4. (a) The variation of E± (B, ) − E0,0 with B reproduces well the experimental observations. For large values of B such that ω ωc , we recover the linear variation with B of the two transition frequencies. When B tends to zero (ωc ω), one ﬁnds two diﬀerent Bohr frequencies corresponding respectively to the two transitions nx = ny = 0 → nx = 0, ny = 1 and nx = ny = 0 → nx = 1, ny = 0. (b) When B tends to zero, one ﬁnds experimentally that the limit of (ν+ − ν− )/(ν+ + ν− ) is of the order of 0.06. The theoretical prediction for this ratio is /2. We therefore conclude that 0.12.

21.7 Comments Quantum boxes of semiconductors, a simple model of which has been examined here, are the subject of many investigations both academic (Coulomb correlations) and applied (optronics). Here we have only considered electronic excitations, but collective modes in a lattice (phonons) also play an important role in the dynamics of quantum boxes. It has been shown recently that the two types of excitations are strongly coupled. This is in contrast with the usual situation encountered in semiconductors, for which the coupling between the electrons and the phonons is weak. The data presented here come from S. Hameau et al., Phys. Rev. Lett. 83, 4152 (1999).

22 Colored Molecular Ions

Some pigments are made of linear molecular ions, along which electrons move freely. We derive here the energy levels of such an electronic system and we show how this energy scheme explains the observed color of the pigments. Consider molecular ions of the chemical formula (Cn Hn+2 )− , which can be considered as deriving from polyethylene molecules, such as hexatriene CH2 =CH-CH=CH-CH=CH2 , with an even number of carbon atoms, by removing a CH+ group. In an ion of this type, the bonds rearrange themselves and lead to a linear structure of the following type: −

(CH2 · · · CH · · · CH · · · CH · · · CH2 )

,

(22.1)

with an odd number n of equally spaced carbon atoms separated by d = 1.4 ˚ A. In this structure, one can consider that the n+1 electrons of the double bonds of the original polyethylene molecule move independently of one another in a one-dimensional inﬁnite potential well of length Ln = nd: V (x) = +∞ =0

for for

x < 0 or x > Ln 0 ≤ x ≤ Ln .

(22.2)

Actually, one should write Ln = (n − 1)d + 2b where b represents the edge eﬀects. Experimentally, the choice b = d/2 appears to be appropriate.

22.1 Hydrocarbon Ions 22.1.1. What are the energy levels εk of an electron in this potential? 22.1.2. Owing to the Pauli principle, at most two electrons can occupy the same energy level. What are the energies of the ground state E0 and of the ﬁrst excited state E 1nof the set of n + 1 electrons? We recall that k=1 k 2 = n(n + 1)(2n + 1)/6.

232

22 Colored Molecular Ions

22.1.3. What is the wavelength λn of the light absorbed in a transition between the ground state and the ﬁrst excited state? One can introduce the A. Compton wavelength of the electron: λC = h/(me c) = 2.426 × 10−2 ˚ 22.1.4. Experimentally, one observes that the ions n = 9, n = 11 and n = 13 A), yellow light (λ11 ∼ 6000 ˚ A) and red light absorb blue light (λ9 ∼ 4700 ˚ A), respectively. Is the previous model in agreement with this (λ13 ∼ 7300 ˚ observation? Are the ions n ≤ 7 or n ≥ 15 colored?

22.2 Nitrogenous Ions One can replace the central CH group by a nitrogen atom, in order to form ions of the type: −

(CH2 · · · CH · · · N · · · CH · · · CH2 )

.

(22.3)

The presence of the nitrogen atom does not change the distances between atoms but it changes the above square well potential. The modiﬁcation consists in adding a small perturbation δV (x), attractive and localized around the nitrogen atom: δV (x) = 0 = −V0

for for

|x − |x −

Ln 2 | Ln 2 |

> α/2 ≤ α/2 ,

where α/d 1 and V0 > 0. 22.2.1. Using ﬁrst order perturbation theory, give the variations δεk of the energy level εk of an electron in the well. For convenience, give the result to leading order in α/d. 22.2.2. Experimentally, one observes that, for the same value of n, the spectrum of the nitrogenous ions (22.3) is similar to that of the ions (22.1) but that the wavelengths λN n are systematically shorter (blue-shifted) if n = 4p+1, and systematically longer (red-shifted) if n = 4p + 3, than those λ0n of the corresponding hydrocarbons (22.1). Explain this phenomenon and show that λN n and λ0n are related by: n+1 λ0n n , = 1 − (−1) 2 γ N λn n+2

where γ is a parameter to be determined. ˚ 22.2.3. The nitrogenous ion n = 11 absorbs red light (λN 11 ∼ 6700 A). Check ˚ ∼ 4300 A ). What is the color of that the ion n = 9 absorbs violet light (λN 9 the nitrogenous ion n = 13? 22.2.4. For suﬃciently large n, if the nitrogen atom is placed not in the central site but on either of the two sites adjacent to the center of the chain,

22.3 Solutions

233

one observes the reverse eﬀect, as compared to question 2.2. There is a red shift for n = 4p + 1 and a blue shift for n = 4p + 3. Can you give a simple explanation for this eﬀect?

22.3 Solutions Section 22.1: Hydrocarbon Ions 22.1.1. The energy levels are εk =

h2 k 2 π2 ¯ 2mL2n

k = 1, 2, . . . .

22.1.2. The ground state energy of the n + 1 electrons is E0 =

h2 π2 ¯ mL2n

(n+1)/2

k2 =

k=1

h2 π2 ¯ (n + 1)(n + 2)(n + 3) . 24 mL2n

The energy of the ﬁrst excited state is E1 = E0 +

# h2 " π2 ¯ π 2 ¯h2 (n + 3)2 − (n + 1)2 = E0 + (n + 2) . 2 8 mLn 2 mL2n

22.1.3. One has hν = E1 − E0 = π 2 ¯ h2 (n + 2)/(2 mL2n ). Since λ = c/ν, we obtain an absorption wavelength λn =

8 d2 n2 . λC (n + 2)

22.1.4. From the general form λn = 646.33 n2 /(n + 2), we obtain λ9 = A, λ13 = 7280 ˚ A, in good agreement with experiment. 4760 ˚ A, λ11 = 6020 ˚ A and λ5 = 2310 ˚ A are in the For smaller n, the wavelengths λ7 = 3520 ˚ ultraviolet part of the spectrum. The ions n ≤ 7 do not absorb visible light and are thus not colored. A and λ17 = 9830 ˚ A are in the For n ≥ 15, the wavelengths λ15 = 8550 ˚ infrared region. These ions do not absorb visible light in transitions from the ground state to the ﬁrst excited state. They are nevertheless colored because of transitions to higher excited states. Section 22.2: Nitrogenous Ions 2/Ln sin(kπx/Ln ). 22.2.1. The normalized wave functions are ψk (x) = One has Ln +α/2 |ψk (x)|2 dx . δεk = δV (x) |ψk (x)|2 dx = −V0 Ln −α/2

234

22 Colored Molecular Ions

Setting y = x − Ln /2, one obtains δεk = −

2V0 Ln

+α/2

sin2 −α/2

kπ kπy + 2 nd

There are two cases: • k even:

2V0 +α/2 2 kπy sin δεk = − dy , Ln −α/2 nd

i.e.

dy .

3

δεk = O((α/d) ) .

The perturbation is negligible. • k odd:

kπy 2V0 +α/2 cos2 δεk = − dy . Ln −α/2 nd To ﬁrst order in α/d, we have δεk = −2V0 α/nd < 0. The exact formulas are: δεk = −

kπα V0 k Ln sin α − (−1) . Ln kπ Ln

The (single particle) energy levels corresponding to even values of k are practically unaﬀected by the perturbation; only those with k odd are shifted. This is simple to understand. For k even, the center of the chain is a node of the wave function, and the integral deﬁning δεk is negligible. For k odd, on the contrary, the center is an antinode, we integrate over a maximum of the wave function, and the perturbation is maximum. 22.2.2. The perturbation to the excitation energy E1 − E0 of question 1.2 is δE = δε(n+3)/2 − δε(n+1)/2 . • (n + 1)/2 even, i.e. n = 4p + 3, δε(n+1)/2 = 0, δE = δε(n+3)/2 = −

2V0 α 0. nd

We can summarize these results in the compact form

n+1 h2 n + 2 π2 ¯ n 2 E1 − E0 + δE = γ 1 − (−1) , 2md2 n2 n+2

22.4 Comments

235

with γ = 4V0 αmd/(π¯ h)2 . We therefore obtain the desired relation n+1 λ0n n . = 1 − (−1) 2 γ N λn n+2

For n = 4p+1, the perturbation increases the excitation energy, and decreases λn . For n = 4p + 3, it decreases the excitation energy, and increases λn . 22.2.3. For the ion n = 11 one obtains the relation (1 − 11γ/13) = 6000/6700, ˚ therefore γ ∼ 0.12 and λN 9 = 4330 A, in good agreement with experiment. One ˚ = 6600 A , which absorbs red light and gives a green color also obtains λN 13 to the corresponding pigment. Note that the presence of the nitrogen atom N 0 0 yields λN 13 ≤ λ11 whereas λ13 > λ11 . 22.2.4. The distance between a node and an antinode of ψk (x) is δx = nd/(2k). For k = (n + 1)/2 and k = (n + 3)/2 which are the states of interest, we will have respectively δx = nd/(n + 1) and δx = nd/(n + 3), i.e. δx ∼ d if n is large. Consequently, if a wave function has a node at the center, it has an antinode in the vicinity of the two adjacent sites, and vice versa. The argument is therefore similar to the answer to questions 2.1 and 2.2, with the reverse eﬀect. The lines are red-shifted if n = 4p + 1 and they are blue-shifted if n = 4p + 3.

22.4 Comments Many further details can be found in the article by John R. Platt, The Chemical Bound and the Distribution of Electrons in Molecules, D Conjugated Chains,, Handbuch Der Physik, Volume XXXVII/2, p. 173, Springer-Verlag (1961). This article is a very complete work on the applications of Quantum Mechanics in Chemistry.

23 Hyperﬁne Structure in Electron Spin Resonance

Many molecular species, such as free radicals, possess an unpaired electron. The magnetic spin resonance of this electron, called electron spin resonance (ESR) as opposed to nuclear magnetic resonance, provides useful information about the electronic structure of the molecule, as we shall see in this chapter. We assume here the following: 1. Spin variables and space variables are independent, both for electrons and for nuclei; we are only interested in the former. 2. The spatial ground state of the unpaired electron is non-degenerate, and one can neglect the eﬀect of a magnetic ﬁeld on its wave function. 3. We only take into account the following magnetic spin interactions: (a) the Zeeman interaction of spin magnetic moments with an external ﬁeld B, and (b) the hyperﬁne interaction between the outer electron and the nuclei. 4. For a given nucleus in the molecule, the hyperﬁne interaction has the form ˆ · Iˆ = (A/4)σ ˆ =h ˆ HF = (A/¯ ˆe · σ ˆ n where S ˆ e /2 is the electron spin h2 ) S ¯σ H ˆ ˆ e and σ ˆ n are the Pauli matrices ˆ n /2 is the nuclear spin; σ and I = h ¯σ which act respectively in the Hilbert spaces of the electron and of the nucleus. The constant A is given by 2 h2 |ψ(r n )|2 , A = − µ0 γe γn ¯ 3 where µ0 = 1/0 c2 is the magnetic susceptibility of vacuum, γe and γn are the gyromagnetic factors of the electron and of the nucleus under consideration, and ψ(r n ) is the value of the electron wave function at the position r n of this nucleus. 5. In all the problem, the system is considered to be in a constant uniform magnetic ﬁeld B directed along the z axis. For simplicity, we set A = h ¯ a, ωe = −γe B, ωn = −γn B and η = (ωe − ωn )/2. The numerical values of gyromagnetic ratios are

238

23 Hyperﬁne Structure in Electron Spin Resonance

electron: proton:

γe /(2π) = −28.024 GHz T−1 , γp /(2π) = +42.574 MHz T−1 .

23.1 Hyperﬁne Interaction with One Nucleus 23.1.1. We ﬁrst consider a species where the nuclei do not possess a magnetic moment, so that there is no hyperﬁne interaction. Write the Zeeman interaction Hamiltonian of the electron with the magnetic ﬁeld B. What are the energy levels of the system? What is the value of the frequency that can excite the system? Give its numerical value for a magnetic ﬁeld of 1 Tesla. 23.1.2. We now assume that the molecule has one spin-1/2 nucleus. We note the (factorized) eigenbasis common to Sˆz and Iˆz as {|σe ; σn } with σe = ±1 and σn = ±1. (a) Write the complete spin Hamiltonian. (b) Calculate the action of σ e · σ n on the vectors of the basis {|σe ; σn }. (c) Write the matrix form of the Hamiltonian in this basis, and calculate its eigenvalues. 23.1.3. From now on, we assume that the magnetic ﬁeld B is strong, in the sense that |ωe | |a|. (a) Give the approximate form of the eigenvalues to ﬁrst order in a/η. (b) Recover these results by ﬁrst diagonalizing the electron Zeeman Hamiltonian, and by treating the other terms, i.e. the nuclear Zeeman Hamiltonian and the hyperﬁne interaction, in ﬁrst order perturbation theory. What are the corresponding eigenstates (to zeroth order in a/η)? (c) One can show that the transitions that an electromagnetic ﬁeld can induce occur only between states which diﬀer by the value of a single spin (for instance, the transitions |+; − → |−; + are forbidden). Under these conditions, what are the observable transition frequencies, knowing that all transitions which are not forbidden actually occur? Classify these transitions in two sets corresponding respectively to nuclear and to electronic spin transitions. (d) Calculate these frequencies numerically for the hydrogen atom in a ﬁeld B = 1 T. We recall that, in this case, A/(2π¯h) 1.420 GHz.

23.2 Hyperﬁne Structure with Several Nuclei We now assume that the molecule has N protons in hydrogen atoms located on sites r 1 , . . . , r N , whose spins are denoted Iˆ 1 , . . . , Iˆ N . The Hilbert space of spin degrees of freedom is of dimension 2N +1 . It is spanned by the set:

23.3 Experimental Results

239

{|σe ; σ1 , σ2 , . . . , σN } ≡ {|σe ⊗ |σ1 ⊗ |σ2 ⊗ · · · ⊗ |σN } with σe = ±1 and σk = ±1, k = 1, . . . , N . This set is an orthonormal eigenbasis common to the z projection of the spin observables Sˆz and Iˆkz , k = 1, . . . , N , of the N + 1 particles. ¯ ak be the hyperﬁne constant of proton k. Write the ex23.2.1. Let Ak = h pression for the spin Hamiltonian of the system (we recall that the magnetic nucleus–nucleus interaction is neglected). 23.2.2. Show that the restriction of this Hamiltonian to each eigen-subspace of Sˆz is diagonal. 23.2.3. Assuming, as in 1.3, that the ﬁeld is strong, calculate the eigenvalues in ﬁrst order perturbation theory, and the corresponding eigenstates. 23.2.4. What are the observable electron spin transition frequencies? How many lines corresponding to these frequencies should the spectrum display in principle? 23.2.5. What is the number of lines and the multiplicity of each of them (i.e. the number of transitions at the same frequency) if all the protons are equivalent, i.e. if all the |ψ(r k )|2 , and therefore the coeﬃcients ak , are equal? 23.2.6. What is the number of lines and their multiplicities, if there exist two sets of equivalent protons, one with p protons corresponding to the constant ap , the other with q = N − p protons, corresponding to the constant aq ?

23.3 Experimental Results Experimentally, one measures the positions and the intensities of the absorption lines in the microwave region. An absorption line appears as a peak in the absorbed intensity α(ν) as a function of the frequency, whose qualitative shape is shown in Fig. 23.1. It can be shown that the intensity of an absorption peak at a given frequency is proportional to the number of transitions (multiplicity of the line) which can occur at that frequency. For experimental convenience, one ﬁxes the frequency of the microwave at a given value, and one varies the magnetic ﬁeld B. This results in an absorption curve α(B). 23.3.1. Figure 23.2 shows the spectrum of the free radical • CH3 (methyl) (J.N. Chazalviel, private communication). The carbon nucleus does not possess any magnetic moment; only the protons of the hydrogen atoms give rise to hyperﬁne interactions. (a) Interpret this spectrum qualitatively. Explain the number of lines and their relative intensities. How many diﬀerent coeﬃcients ak are there?

240

23 Hyperﬁne Structure in Electron Spin Resonance

Fig. 23.1. Typical shape of an ESR absorption curve as a function of the frequency

2,3 10-3 T

α (B) B Fig. 23.2. Microwave spectrum of the radical • CH3

(b) Give the value of ak /(2π). Calculate the value of |ψ(r k )|2 for the unpaired electron in this molecule. It is convenient to express the result in terms of |ψ(0)|2Hydrogen = 1/(πa31 ) where a1 is the Bohr radius of hydrogen. 23.3.2. Answer the same questions for the spectrum of CH3 −• COH−COO− (the radical ion of lactic acid) shown in Fig. 23.3. Neither the oxygen nor the carbon nuclei carry magnetic moments. The only hyperﬁne interaction arises, again, from the protons of the hydrogen atoms.

23.4 Solutions Section 23.1: Hyperﬁne Interaction with One Nucleus ˆ = −¯ ˆez /2, hence the energy 23.1.1. The magnetic Hamiltonian is H hγe B σ hγe B/2 corresponding to the states |± . The transition levels E± = ∓¯ ¯ ωe , ν = ωe /(2π). For B = 1 T, frequency is given by hν = E+ − E− = h ν = 28.024 GHz. 23.1.2. (a) The full Hamiltonian, including the hyperﬁne interaction, is

23.4 Solutions

241

1,7110-3 T αB

2 10-4 T

B

Fig. 23.3. Microwave spectrum of the radical CH3 −• COH−COO−

ˆ · Iˆ ˆ = −γe B Sˆz − γn B Iˆz + A S H h2 ¯ hωn ¯ ¯ha hωe ¯ ˆn . ˆe · σ σ ˆez + σ ˆnz + σ = 2 2 4 ˆ n on the basis states is: ˆe · σ (b) The action of σ ˆe · σ ˆ n |+; + = |+; + σ ˆe · σ ˆ n |+; − = 2|−; + − |+; − σ ˆe · σ ˆ n |−; + = 2|+; − − |−; + σ ˆ n |−; − = |−; − . ˆe · σ σ (c) Hence the 4 × 4 matrix representation of the Hamiltonian ⎛ ⎞ 0 0 a + 2(ωe + ωn ) 0 ⎟ h⎜ 0 4η − a 2a 0 ˆ =¯ ⎟ , ⎜ H ⎠ ⎝ 0 2a −4η − a 0 4 0 0 0 a − 2(ωe + ωn ) where the rows and columns are ordered as |+; +, |+; −, |−; +, |−; −. Hence the eigenstates and the corresponding eigenvalues: ¯ h (a + 2(ωe + ωn )) 4 h ¯ |−; − −→ (a − 2(ωe + ωn )) 4 |+; + −→

and from the diagonalization of the 2 × 2 matrix between |+; − and |−; + ¯ h (−a + 2 4η 2 + a2 ) 4 h ¯ sin φ |+; − − cos φ |−; + −→ (−a − 2 4η 2 + a2 ) 4

cos φ |+; − + sin φ |−; + −→

with tan φ =

a . 2η + 4η 2 + a2

242

23 Hyperﬁne Structure in Electron Spin Resonance

23.1.3. (a) If η a, the eigenvectors and eigenvalues are, to lowest order, |+; + |+; − |−; + |−; −

−→ (¯ h/4)(a + 2(ωe + ωn )) −→ ∼ (¯ h/4)(4η − a) −→ ∼ (¯ h/4)(−4η − a) −→ (¯ h/4)(a − 2(ωe + ωn )) .

(b) In each subspace corresponding respectively to σe = 1 and σe = −1, the perturbation is diagonal (the non-diagonal terms couple σe = +1 and σe = −1). The 2 × 2 matrices to be considered are indeed ˆ σn +, σn |H|+,

ˆ and −, σn |H|−, σn .

ˆ Consider for instance +, σn |H|+, σn . Since +, σn |Sˆx |+, σn = +, σn |Sˆy |+, σn = 0 , only +|Sˆz |+σn |Iˆz |σn has to be considered, and it is diagonal. The eigenstates at zeroth order are therefore |σe ; σn and we recover the above results. (c) Transitions: (i) Nuclear transitions: |σe ; + ↔ |σe ; − , i.e. |+; + ↔ |+; − |−; + ↔ |−; −

∆E = h ¯ (ωn + a/2), ∆E = h ¯ (ωn − a/2),

ν = |ωn + a/2|/(2π) ν = |ωn − a/2|/(2π) .

(ii) Electronic transitions: |+; σn ↔ |−; σn , i.e. |+; + ↔ |−; + |+; − ↔ |−; −

∆E = h ¯ (ωe + a/2), ∆E = h ¯ (ωe − a/2),

ν = |ωe + a/2|/(2π) ν = |ωe − a/2|/(2π) .

(d) For B = 1 T, νn = 42.6 MHz; a/(2π) = A/(2π¯h) = 1420 MHz; νe = 28.024 GHz. The nuclear transitions occur at ν1 = 753 MHz and ν2 = 667 MHz, the electronic transitions occur at ν1 = 28.734 GHz and ν2 = 27.314 GHz. Section 23.2: Hyperﬁne Structure with Several Nuclei 23.2.1. The total Hamiltonian is N

N

k=1

k=1

¯ Ak hωn hωe ˆ =¯ ˆe · σ ˆk . σ ˆez + σ ˆkz + σ H 2 2 4 ˆ to a subspace corresponding to the eigenvalue 23.2.2. The restriction of H ¯ σe /2 of Sˆez (σe = ±) can be written using 1.2(b) or (c): h N

hωn ¯ Ak σe hωe ˆσ = ¯ H σ + + σ ˆkz . e e 2 2 4 k=1

ˆ + and H ˆ − are diagonal in the basis {|σ1 , σ2 , . . . , σN }. The operators H

23.4 Solutions

243

23.2.3. First order Nperturbation theory N consists in diagonalizing the perturbˆ e ·σ ˆ k in each eigen-subspace hωn /2)ˆ σkz + k=1 (Ak /4)σ ing Hamiltonian k=1 (¯ ˆez /2. This is automatically satisﬁed. Therefore, of the dominant term h ¯ ωe σ σe = +1 : Eσ+1 ...σN =

h(2ωn + ak ) hωe ¯ ¯ + σk , state |+; σ1 , . . . , σN , 2 4 k

σe = −1 : Eσ−1 ...σN = −

h(2ωn − ak ) hωe ¯ ¯ + σk , state |−; σ1 , . . . , σN . 2 4 k

23.2.4. There are 2N transitions |+; σ1 , . . . , σN ↔ |−; σ1 , . . . , σN corresponding to the 2N possible choices for the set {σk }. The corresponding frequencies are 1 ∆νσ1 ...σN = ak σk /2 . ω e + 2π k

23.2.5. If all ak are equal to a, we have 1 1 |ωe + M a/2| , σk /2 = ∆ν = ω e + a 2π 2π k

with M = σk = N, N −2, . . . , −N +2, −N , i.e. N +1 absorption lines. There (N −M )/2 are CN transitions which have the same frequency and contribute to each line. The relative intensities of the lines will therefore be proportional to (N −M )/2 . The splitting between two adjacent lines the binomial coeﬃcients CN is a. 23.2.6. If p equivalent protons correspond to the coupling constant Ap , and q = N − p correspond to Aq , then p q 1 1 ap aq ap aq = ω ∆ν = + M + σ + σ + M ω . e i j e p q 2π 2 i=1 2 j=1 2π 2 2 There are p + 1 values of Mp : p, p − 2, . . . , −p, and q + 1 values of Mq : Mq = q, q −2, . . . , −q. The total number of lines is (p+1) (q +1), and the multiplicity (p−Mp )/2 (q−Mq )/2 Cq . of a line corresponding to a given couple (Mp , Mq ) is Cp Section 23.3: Experimental Results 23.3.1. The experimental results conﬁrm the above analysis. 1. For • CH3 there are 4 equally spaced lines of relative intensities 1 : 3 : 3 : 1. This is in perfect agreement with the fact that the three protons of • CH3 are obviously equivalent. All the Ak coeﬃcients are equal.

244

23 Hyperﬁne Structure in Electron Spin Resonance

2. For a ﬁxed ω, one gets by considering two consecutive lines, for instance the center lines: a/2 − γe B1 = −a/2 − γe B2 so that a = γe (B1 − B2 ). We h, and deduce ν = |a|/2π = 65 MHz = |Ak |/2π¯ πa31 |ψ(rk )|2 = |ψ(rk )|2 /|ψ(0)|2Hydrogen = 65/1420 ∼ 0.045 . In the radical • CH3 , the probability that the outer electron is on top of a proton is smaller by a factor 3 × 0.045 = 0.135 than in the hydrogen atom. 23.3.2. In the case of CH3 −• COH−COO− , there are four dominant lines, each of which is split into two. This agrees with the fact that, in the molecule CH3 −• COH−COO− , the 3 protons of the CH3 group are equivalent and have the same hyperﬁne constant a1 whereas the proton of the • COH group has a diﬀerent constant a2 which is noticeably smaller than a1 . A calculation similar to the previous one gives |ψ(rk )|2 /|ψ(0)|2Hydrogen ∼ 0.034 for the protons of the CH3 group, and |ψ(rk )|2 /|ψ(0)|2Hydrogen ∼ 0.004 for the proton of • COH.

24 Probing Matter with Positive Muons

A very eﬃcient technique for probing the structure of crystals consists in forming, inside the material, pseudo hydrogenic atoms made of an electron and a positive muon, and called muonium. This chapter is devoted to the study of the dynamics of muonium, both in vacuum and in a silicon crystal. The positive muon is a spin-1/2 particle which has the same charge as the proton. The muon mass is considerably larger than the electron mass: mµ /me = 206.77. The muon is unstable and decays with a lifetime τ = 2.2 µs. Its use in probing the structure of crystals is based on the rotation of its spin, once a muonium atom is formed: •

It is possible to form muonium atoms in a quantum state such that, at t = 0, the spin state of the µ+ is known. • Using a technique of particle physics, one can measure its spin state at a later time t. • The rotation of the muon spin can be related to the hyperﬁne structure of the 1s level of muonium.

Therefore, the muonium constitutes a local probe, sensitive to electric and magnetic ﬁelds in its vicinity. One can obtain in this way information on the structure of the medium by methods analogous to magnetic resonance experiments. In the ﬁrst part of the chapter, we sketch the principle of the method by studying muonium in vacuum. When the method was ﬁrst applied to a silicon crystal, in 1973, the results seemed anomalous. We shall see in the second section how these results were understood, in 1978, as being due to the anisotropy of crystalline media. Throughout this chapter, the muon will be considered as stable. For simplicity, we set ˆ1 ˆe ≡ µ ˆ2 , ˆ 1 = µ1 σ ˆ 2 = µ2 σ ˆ µ+ ≡ µ µ µ ˆ 1 and σ ˆ 2 are the Pauli matrices. where the (x, y, z) components of σ Numerical values of interest are:

246

24 Probing Matter with Positive Muons

mµ c2 = 105.66 MeV me c2 = 0.511 MeV

µ1 /h = 67.5 MHz T−1 µ2 /h = −1.40 × 104 MHz T−1 .

24.1 Muonium in Vacuum Muonium is formed by slowing down a beam of µ+ , prepared in a given spin state, in a thin metal foil. A suﬃciently slow µ+ can capture an electron and form a hydrogen-like atom in an excited state. This atom falls into its ground state very quickly (in ∼ 10−9 s), the muon’s spin state remaining the same during this process. Once it is formed, the muonium, which is electrically neutral, can diﬀuse outside the metal. We assume that, at t = 0, the state of the muonium atom is the following: • • •

The muon spin is in the eigenstate | + z ≡ |+ of σ ˆ1z . The electron spin is in an arbitrary state α|++β|−, with |α|2 +|β|2 = 1. The wave fuction Ψ (r) of the system is the 1s wave function of the hydrogen-like system, ψ100 (r).

Just as for the hyperﬁne structure of hydrogen, we work in the 4 dimensional Hilbert space corresponding to the spin variables of the electron and the muon. In this Hilbert space, the spin–spin interaction Hamiltonian is A ˆ = E0 − 2 µ0 |ψ100 (0)|2 µ ˆ 2 = E0 + σ ˆ2 , ˆ1 · µ ˆ1 · σ H 3 4π 4 where the indices 1 and 2 refer respectively to the muon and to the electron, and where E0 = −mr c2 α2 /2, with mr being the reduced mass of the (e, µ) system. ˆ in the basis 24.1.1. Write the matrix representation of the Hamiltonian H {|σ1z , σ2z , σiz = ±}. 24.1.2. Knowing the value of A in the hydrogen atom: A/h = 1420 MHz, calculate A in muonium. We recall that µ1 = q¯h/(2mµ ) for the muon, µ2 = h/(2mp ) for the proton, where q is −q¯ h/(2me ) for the electron, µp = 2.79 q¯ the unit charge and mp = 1836.1 me . 24.1.3. Write the general form of an eigenstate of σ ˆ1z with eigenvalue +1: ˆ (i) in the basis {|σ1z , σ2z }; (ii) in the eigenbasis of H. 24.1.4. We assume that, at t = 0, the system is in a state |ψ(0) of the type deﬁned above. Calculate |ψ(t) at a later time. 24.1.5. (a) Show that the operators π ˆ± = (1 ± σ ˆ1z )/2 are the projectors on the eigenstates of σ ˆ1z corresponding to the eigenvalues ±1. (b) Calculate for the state |ψ(t) the probability p(t) that the muon spin is in the state |+ at time t. Write the result in the form

24.2 Muonium in Silicon

247

p(t) = q p+ (t) + (1 − q) p− (t) , where p+ (or p− ) is the probability that one obtains if the electron is initially in the eigenstate of σ2z with eigenvalue +1 (or −1), and where q is a probability, as yet undeﬁned. 24.1.6. In practice, the electronic spins are unpolarized. A rigorous treatment of the problem then requires a statistical description in terms of a density operator. To account for this nonpolarization in a simpler way, we shall set heuristically that the observed probability p¯(t) corresponds to q = 1/2 in the above formula. Using this prescription, give the complete expression for p¯(t).

24.2 Muonium in Silicon We now form muonium in a silicon crystal suﬃciently thick that the muonium does not escape. The muonium stops in an interstitial position inside the crystal lattice, the nearest atoms forming a plane hexagonal mesh around it. The global eﬀect of the interactions between the atoms of the crystal and the muonium atom is to break the spherical symmetry of the spin–spin interaction, but to preserve the rotational symmetry around the z axis perpendicular to the plane of the mesh. We therefore consider the Hamiltonian ˆ = E0 + A σ ˆ2 + D σ ˆ1 · σ ˆ1z σ ˆ2z , H 4

where the constant A may diﬀer from A since the presence of neighboring atoms modiﬁes the Coulomb potential and, therefore, the wave function at the origin. The constants A and D will be determined experimentally; their sign is known: A > 0, D < 0. 24.2.1. Calculate the spin energy levels and the corresponding eigenstates of the muonium trapped in the silicon crystal. 24.2.2. We now reconsider the spin rotation experiment with the following modiﬁcations: • Initially the µ+ spin is now in the eigenstate | + x of σx . • We want to know the probability of ﬁnding the µ+ spin in this same eigenstate | + x at time t. One can proceed as in question 1.5: (a) Calculate in the {|σ1 , σ2 } basis the states |ψ+ (t) and |ψ− (t) which are σ2z is the projection of the electron spin along initially eigenstates of σ ˆ2z (ˆ the z axis). σ1x |ψ (t), where = ±. (b) Evaluate ψ (t)|ˆ

248

24 Probing Matter with Positive Muons

(c) Consider the projector π ˆx = (1 + σ ˆ1x )/2 , and deduce from the previous question the probabilities p± (t). (d) Calculate the measured probability p¯(t) = (p+ (t) + p− (t))/2. 24.2.3. Comparison with experiment: Present day technology in data processing allows one to determine not p(t) itself, but a quantity which is easier to deal with, the characteristic function g(ω) = Re(f (ω)) where ∞ 1 f (ω) = p¯(t) e−t/τ eiωt dt τ 0 is the Fourier transform of p¯(t)e−t/τ /τ . In this expression, the factor e−t/τ /τ is due to the ﬁnite lifetime of the µ+ (τ ∼ 2.2 µs). We recall that ∞ 1 1 . e−t/τ eiωt dt = τ 0 1 − iωτ (a) Figure 24.1a shows the distribution g(ω) as measured in the conditions of question 2.2. Check that this data is compatible with the results found in question 2.2, and deduce from the data the values of A /h and D/h (we recall that D < 0). g(ω)

37,25

54,85

(arbitrary units) (a)

0

10

100

50

(

)

(

)

ω/2π MHz

g(ω) (b) 92,1

0

10

50

100

ω/2π MHz

Fig. 24.1. Experimental variations of the quantity g(ω), deﬁned in the text, with the frequency ν = ω/(2π). (a) In the conditions described in question 2.2, and (b) in another experimental conﬁguration

24.3 Solutions

249

(b) Figure 24.1b is obtained by a slight modiﬁcation of the previous experiment. Can you tell what modiﬁcation has been made? How can one evaluate the position of the third peak, in terms of the constants of the problem?

24.3 Solutions Section 24.1: Muonium in Vacuum 24.1.1. The Hamiltonian is ˆ = E0 + A (ˆ H σ1x σ ˆ2x + σ ˆ1y σ ˆ2y + σ ˆ1z σ ˆ2z ) . 4 The matrix representation is therefore ⎛ ⎞ E0 + A/4 0 0 0 ⎜ ⎟ A/2 0 0 E0 − A/4 ˆ =⎜ ⎟ , H ⎝ ⎠ 0 0 A/2 E0 − A/4 0 0 0 E0 + A/4 where the elements are ordered as: | + +, | + −, | − +, | − −. 24.1.2. The constant A is related to its value in the hydrogen atom by A |ψ(0)|2 µ1 |ψ(0)|2 mp 1 . = = 2 AH |ψ(0)|H µp |ψ(0)|2H mµ 2.79 In ﬁrst approximation, muonium and hydrogen have similar sizes and wave functions, since the muon is much heavier than the electron. Therefore we obtain A AH (mp /2.79 mµ ) and A/h 4519 MHz. The reduced mass correction to the value of the wave function at the origin is straightforward to calculate. It is of the order of 1% and it leads to A = 4519 (1 − 0.0126) = 4462 MHz . h This value is very close to the observed 4463 MHz, the diﬀerence being due to relativistic eﬀects. 24.1.3. The state under consideration can be written as |ψ = |+ ⊗ (α|+ + β|−)

with |α|2 + |β|2 = 1 .

Equivalently, one can write it as |ψ = α| + + + β| + −. ˆ consists in the common eigenstates of the total spin The eigenbasis of H 2 ˆ ˆ operators S and Sz : ⎧ ⎨ | + + √ triplet states (| + − + | − +)/ 2 ⎩ | − − √ singlet state (| + − − | − +)/ 2 .

250

24 Probing Matter with Positive Muons

Therefore, one also has the representation β |ψ = α|1, 1 + √ (|1, 0 + |0, 0) , 2 where the only constraint on α and β is |α|2 + |β|2 = 1. 24.1.4. We start from |ψ(0) = |ψ as deﬁned above. The energy levels and the corresponding eigenstates are known: triplet states

ET = E0 + A/4

singlet state

ES = E0 − 3A/4 .

At time t the state is:

β i3At/4¯h β −iE0 t/¯ h −iAt/4¯ h |0, 0 . |ψ(t) = e e α |1, 1 + √ |1, 0 + √ e 2 2 24.1.5. (a) It is straightforward to check that π ˆ± are projectors: ˆ+ |− = 0 π ˆ+ |+ = |+ π ˆ− |+ = 0 . π ˆ− |− = |− π (b) The probability of ﬁnding the muon spin in the state |+ at time t is by deﬁnition p(t) = ˆ π+ |ψ(t)2 = ψ(t)|ˆ π+ |ψ(t) . Using π ˆ+ |1, 1 = |1, 1 1 ˆ+ |0, 0 = √ | + − π ˆ+ |1, 0 = π 2 π ˆ+ |1, −1 = 0 , we obtain β π ˆ+ |ψ(t) = e−i(E0 +A/4)t/¯h α| + + + 1 + eiAt/¯h | + − . 2 Squaring the norm of π ˆ+ |ψ(t), we get h)) . p(t) = |α|2 + |β|2 cos2 (At/(2¯ There is a periodic modulation of the probability of observing the muon spin aligned with the positive z axis, which can be interpreted as a rotation of the muon spin with frequency ν = A/h.

24.3 Solutions

251

• The probability p+ (t) corresponds to the initial state |ψ(0) = |++. This is a stationary state so that p(t) ≡ p+ (t) = 1 in this case. to the initial state |ψ(0) = • The probability p− (t) corresponds √ | + − = (|1, 0 + |0, 0)/ 2. There is in this case an oscillation with a 100 % modulation between | + − and | − +, so that p(t) ≡ p− (t) = h)). cos2 (At/2(¯ Therefore the result can be cast in the form suggested in the text: p(t) = q p+ (t) + (1 − q) p− (t) , with q = |α|2 . 24.1.6. When the electronic spins are unpolarized, we obtain following the assumption of the text: p¯(t) =

3 1 + cos(At/¯h) . 4 4

Note: The rigorous way to treat partially polarized systems is based on the density operator formalism. In the present case the density operator for the unpolarized electron is: ρ2 =

1 (|++| + |−−|) , 2

so that the initial density operator for the muonium is: 1 1 | + ++ + | + | + −+ − | 2 2 1 = |1, 11, 1| 2 1 + (|1, 01, 0| + |1, 00, 0| + |0, 01, 0| + |0, 00, 0|) . 4

ρ(0) =

The density operator at time t is then given by: ρ(t) =

1 |1, 11, 1| 2 1 + |1, 01, 0| + e−iAt/¯h |1, 00, 0| 4 +eiAt/¯h |0, 01, 0| + |0, 00, 0|

hence the probability: p¯(t) = +, +|ρ(t)|+, + + + − |ρ(t)| + −

1 1 1 1 1 1 = + + e−iAt/¯h + eiAt/¯h + 2 4 2 2 2 2 3 1 h) . = + cos(At/¯ 4 4

252

24 Probing Matter with Positive Muons

Section 24.2: Muonium in Silicon 24.2.1. In the factorized basis {|σ1 , σ2 }, the Hamiltonian is written as ⎞ ⎛ 0 0 0 A /4 + D ⎟ ⎜ A /2 0 0 −A /4 − D ˆ = E0 + ⎜ ⎟ . H ⎠ ⎝ −A /4 − D 0 0 A /2 0 0 0 A /4 + D This Hamiltonian is diagonal in the eigenbasis {|S, m} of the total spin. A simple calculation shows that the eigenvalues and eigenvectors are E1 = E4 = E0 + A /4 + D E2 = E0 + A /4 − D E3 = E0 − 3A /4 − D

|1, 1 and |1, −1 |1, 0 |0, 0 .

24.2.2. (a) The initial states |ψ+ (0) and |ψ− (0) are easily obtained in the factorized basis as √ |ψ+ (0) = | + x ⊗ |+ = (| + + + | − +)/√2 |ψ− (0) = | + x ⊗ |− = (| + − + | − −)/ 2 . They can be written in the total spin basis {|S, m} as 1 1 |ψ+ (0) = √ |1, 1 + (|1, 0 − |0, 0) 2 2 1 1 |ψ− (0) = √ |1, −1 + (|1, 0 + |0, 0) . 2 2 h, we ﬁnd at time t: Writing ωi = −Ei /¯ eiω3 t eiω2 t eiω1 t |ψ+ (0) = √ |1, 1 + |1, 0 − |0, 0 2 2 2 eiω3 t eiω2 t eiω4 t |ψ− (0) = √ |1, −1 + |1, 0 + |0, 0 , 2 2 2 which can now be converted in the factorized basis: eiω2 t − eiω3 t eiω2 t + eiω3 t eiω1 t √ √ |ψ+ (t) = √ | + + + | + − + | − + 2 2 2 2 2 eiω2 t + eiω3 t eiω2 t − eiω3 t eiω4 t √ √ |ψ− (t) = √ | − − + | + − + | − + . 2 2 2 2 2

24.3 Solutions

253

(b) Since σ ˆ1x |σ1 , σ2 = | − σ1 , σ2 , the matrix elements ψ± (t)|ˆ σ1x |ψ± (t) are equal to:

1 Re e−iω1 t eiω2 t + eiω3 t 2

(A + 2D)t 1 2Dt + cos = cos 2 h ¯ ¯h " −iω4 t iω2 t

# 1 e σ1x |ψ− (t) = Re e + eiω3 t . ψ− (t)|ˆ 2 ψ+ (t)|ˆ σ1x |ψ+ (t) =

Since ω1 = ω4 , the two quantities are equal. (c) The desired probabilities are p± (t) = ˆ π+x |ψ± (t)2 = ψ± (t)|ˆ π+x |ψ± (t) or, equivalently, 1 1 1 ˆ1x ) |ψ± (t) = + ψ± (t)|ˆ p± (t) = ψ± (t)| (1 + σ σ1x |ψ± (t) . 2 2 2 Using the result obtained above, we get:

(A + 2D)t 1 1 2Dt + cos p± (t) = + cos . 2 4 h ¯ ¯h (d) Since p+ (t) = p− (t), the result for p¯(t) is simply:

(A + 2D)t 1 1 2Dt + cos p¯(t) = + cos . 2 4 h ¯ ¯h 24.2.3. Comparison with Experiment: In practice, the time t corresponds to the decay of the µ+ , with the emission of a positron e+ and two neutrinos. The positron is suﬃciently energetic and leaves the crystal. It is emitted preferentially in the muon spin direction. One therefore measures the direction where the positron is emitted as a function of time. For N0 incoming muons, the number of positrons emitted in the x direction is dN (t) = N0 p¯(t)e−t/τ dt/τ , where τ is the muon lifetime. A convenient way to analyse the signal, and to extract the desired frequencies, consists in taking the Fourier transform of the above signal. Deﬁning ∞ 1 1 , e(iω−1/τ )t dt = f0 (ω) = τ 0 1 − iωτ one obtains

1 1 2D 2D f (ω) = f0 (ω) + f0 ω − + f0 ω + 2 8 h ¯ ¯h

1 A + 2D A + 2D + f0 ω − + f0 ω + . 8 h ¯ ¯h

254

24 Probing Matter with Positive Muons

The function Re(f0 (ω)) has a peak at ω = 0 whose half width is 1/τ , which corresponds to 100 kHz. (a) The curve of Fig. 24.1 is consistent with this observation. Besides the h. peak at ω = 0, we ﬁnd two peaks at ω1 = −2D/¯h and ω2 = (A + 2D)/¯ Assuming that D is negative, which can be conﬁrmed by a more thorough analysis, one obtains 2D/h = −37.25 MHz

and

A /h = 92.1 MHz .

(b) In general, one expects to see peaks at all frequencies ωi − ωj , and in h. In order to observe the corresponding peak, one particular at ω2 −ω3 = −A /¯ must measure the µ+ spin projection along a direction which is not orthogonal to the z axis. This leads to a term in cos (ω2 − ω3 )t in p¯(t), which appears in Fig. 24.1.

25 Quantum Reﬂection of Atoms from a Surface

This chapter deals with the reﬂection of very slow hydrogen atoms from a surface of liquid helium. In particular, we estimate the sticking probability of the atoms onto the surface. This sticking proceeds via the excitation of a surface wave, called a ripplon. We show that this probability must vanish at low temperatures, and that, in this limit, the reﬂection of the atoms on the surface is specular. In all the chapter, the position of a particle is deﬁned by its coordinates r = (x, y) in a horizontal plane, and its altitude z. The altitude z = 0 represents the position of the surface of the liquid He bath at rest. The wave functions ψ(r, z) of the H atoms are normalized in a rectangular box of volume Lx Ly Lz . We write m for the mass of a H atom (m = 1.67 10−27 kg).

25.1 The Hydrogen Atom–Liquid Helium Interaction Consider a H atom above a liquid He bath at rest (cf Fig. 25.1). We model the H-liquid He interaction as the sum of pairwise interactions between the H atom at point (R, Z) (Z > 0), and the He atoms at (r, z), with z < 0: +∞ dz U ( (R − r)2 + (Z − z)2 ) Θ(−z) V0 (Z) = n d2 r −∞

where n is the number of He atoms per unit volume, and Θ is the Heaviside function. 25.1.1. We recall the form of the Van der Waals potential: U (d) = −

C6 , d6

which describes the long distance interaction between a H atom and a He atom separated by a distance d. Show that the long distance potential between the H atom and the liquid He bath is of the form:

256

25 Quantum Reﬂection of Atoms from a Surface

Fig. 25.1. A hydrogen atom incident on a liquid helium bath. The oscillations of the surface of the liquid are studied in Sect. 25.2

V0 (Z) = −

α . Z3

Express α in terms of C6 and n. 25.1.2. Experimentally, one ﬁnds α = 1.9 × 10−2 eV ˚ A3 . At what distance from the surface does gravity become larger than the Van der Waals force? In what follows, we shall neglect the gravitational force. 25.1.3. Show that the eigenstates of the Hamiltonian which describes the motion of the H atom are of the form |k⊥ , φσ , where k⊥ represents a plane running wave propagating in the plane Oxy, i.e. parallel to the surface of the liquid He, and where φσ is an eigenstate of the Hamiltonian which describes the motion along the z axis: R, Z|k⊥ , φσ =

1 ei(kx X+ky Y ) φσ (Z) . Lx Ly

25.1.4. We want to evaluate the number of bound states of the motion along the z axis in the potential: α Z3 V0 (Z) = +∞ V0 (Z) = −

if Z > zmin if Z ≤ zmin .

We shall use the WKB approximation. (a) Justify the shape of this potential. (b) What is the continuity condition for the wave function at Z = zmin ? (c) Show that the quantization condition for a motion with turning points zmin and b is

b 3 k(Z) dZ = n + π 4 zmin with n integer ≥ 0.

25.2 Excitations on the Surface of Liquid Helium

257

(d) Infer the order of magnitude of the number of bound states as a function of zmin and α. What is the domain of validity of the result? A. (e) The parameter zmin for the surface of liquid He is of the order of 2 ˚ How many bound states does one expect for the motion along the z axis? (f) Experimentally, one ﬁnds that there is a single bound state H–liquid He, whose energy is E0 = −8.6 × 10−5 eV. Compare this result with the WKB prediction. This unique bound state in the z-axis motion will be denoted φ0 in the rest of the chapter.

25.2 Excitations on the Surface of Liquid Helium The general dispersion relation for waves propagating on the surface of a liquid is A ωq2 = gq + q 3 with q = |q| , ρ0 where ωq and q = (qx , qy ) are, respectively, the frequency and the wave vector of the surface wave, g is the acceleration of gravity and A and ρ0 represent the surface tension and the mass density of the liquid. 25.2.1. Discuss the nature of the surface waves (capillary waves or gravity waves) according to the value of the wavelength λ = 2π/q. Perform the numerical application in the case of liquid He: ρ0 = 145 kg m−3 , A = 3.5 × 10−4 Jm−2 . 25.2.2. Hereafter, we are only interested in waves for which h ¯ ωq |E0 |. Show that these are always capillary waves and give their wavelengths. In what follows we shall use the simpler dispersion relation ωq2 = (A/ρ0 )q 3 . 25.2.3. In order to quantize these surface waves, we introduce the bosonic operators rq and rq† corresponding to the annihilation and the creation of an excitation quantum. These elementary excitations are called ripplons. The Hamiltonian which describes these excitations is:1 HS =

q max q

hωq rˆq† rˆq . ¯

The altitude h(r) of the liquid surface at point r = (x, y) becomes a twodimensional scalar ﬁeld operator: ˆ h(r) =

q max q

1

hq (rq† e−iq.r + rq eiq.r )

with

hq =

¯q h . 2ρ0 ωq Lx Ly

The summation over q is limited to q < qmax where qmax is of the order of a fraction of an inverse ˚ Angstrom. For larger values of q, hence smaller wavelengths, the description of the vicinity of the surface in terms of a ﬂuid does not hold any longer.

258

25 Quantum Reﬂection of Atoms from a Surface

Evaluate, at zero temperature, the r.m.s. altitude ∆h of the position of the surface. We recall that, in two dimensions, the conversion of a discrete summation into an integral proceeds via: Lx Ly −→ d2 q . 2 4π q A−1 . Numerical application: qmax = 0.5 ˚

25.3 Quantum Interaction Between H and Liquid He We now investigate the modiﬁcations to the H–liquid He potential arising from the possible motion of the surface of the liquid helium bath. In order to do so, we replace the coupling considered above by +∞ ˆ dz U ( (R − r)2 + (Z − z)2 ) Θ(h(r) − z) . V (R, Z) = n d2 r −∞

ˆ and interpret the result. 25.3.1. Expand V (R, Z) to ﬁrst order in h ˆ 25.3.2. Replacing h(r) by its expansion in terms of operators rˆq , rˆq† , cast V (R, Z) in the form:

hq e−iq.R Vq (Z)rq† + h.c. V (R, Z) = V0 (Z) + q

with Vq (Z) = n

d2 r e−iq.r U ( r 2 + Z 2 ).

25.3.3. Introducing the creation operators a ˆ†k,σ and the annihilation operators a ˆk,σ of a hydrogen atom in an eigenstate of the motion in the potential V0 (Z), write in second quantization the total hydrogen–ripplon Hamiltonian ˆ to ﬁrst order in h.

25.4 The Sticking Probability We consider a H atom in an asymptotically free state in the z direction (i.e. behaving as e±ikσ z as z → +∞). This state denoted |k⊥ , φσ has an energy Ei =

¯2 2 h (k + kσ2 ) . 2m ⊥

We now calculate the probability that this atom sticks on the surface, which is assumed here at zero temperature.

25.5 Solutions

259

25.4.1. How does the matrix element φ0 |Vq |φσ vary with the size of the normalization box? We assume in the following that this matrix element is proportional to kσ if kσ is suﬃciently small, and we introduce M (q) such that φ0 |Vq |φσ = √

hkσ ¯ M (q) . 2mLz

All following results will be expressed in terms of M (q). 25.4.2. Using Fermi’s Golden Rule, deﬁne a probability per unit time for an atom to stick on the surface. In order to do so, one will deﬁne properly: (a) the continuum of ﬁnal states; (b) the conditions imposed by energy conservation. For simplicity, we shall assume that the incident energy Ei is negligible compared to the bound state energy E0 . Show that the emitted ripplon has a wave vector q such that |q| = q0 with:

h ¯

A 3/2 h ¯ 2 q02 = |E0 | ; q0 + ρ0 r2m

(c) the density of ﬁnal states. 25.4.3. Express the ﬂux of incident atoms in terms of ¯h, kσ , m and Lz . 25.4.4. Write the expression for the probability that the hydrogen atom sticks on the surface of the liquid helium bath in terms of ¯h, q0 , A, ρ0 , kσ and M (q). Check that this probability is independent of the normalization volume Lx Ly Lz . 25.4.5. How does this probability vary with the energy of the incident hydrogen atoms? 25.4.6. Describe qualitatively how one should modify the above treatment if the liquid helium bath is not at zero temperature.

25.5 Solutions Section 25.1: The Hydrogen Atom–Liquid Helium Interaction 25.1.1. We use cylindrical coordinates, assuming that the H atom is at R = 0. The potential V0 (Z) takes the form

260

25 Quantum Reﬂection of Atoms from a Surface

V0 (Z) = n

0

d2 r −∞

= −nC6

dz U ( r2 + (Z − z)2 )

0

∞

dz −∞ 0

π = − nC6 2

dr 0

−∞

2πr (r2 + (Z − z)2 )3

dz πnC6 =− . 4 (Z − z) 6Z 3

Therefore

α πnC6 . with α= Z3 6 25.1.2. The force which derives from V0 (Z) has modulus V0 (Z) = −

F (Z) =

3α . Z4

We have 3α/Zg4 = M g for Zg = (3α/(M g))1/4 . The numerical application yields Zg = 0.86 µm which is very large on the atomic scale. For all the relevant H–liquid He distances, which are between 0.1 nm and 1 nm, gravity can be neglected. ˆ Z , where ˆ =H ˆ⊥ + H 25.1.3. The Hamiltonian can be split as H 2 pˆ2 ˆ ⊥ = pˆx + y H 2m 2m

2 ˆ Z = pˆz + V0 (Z) ˆ . H 2m

and

These two Hamiltonians commute and the eigenbasis of the total Hamiltonian ˆ ⊥ , where k⊥ ˆ is factorized as a product |k⊥ , φσ of (i) the eigenstates of H H represents the wave vector of a plane running wave propagating in the (x, y) ˆ Z which describes the motion along the plane, and (ii) the eigenstates φσ of H z axis. 25.1.4. (a) For Z ≤ zmin , the overlap of the electron wave functions of the H and He atoms causes a repulsion between these atoms, which is modeled here by a hard core potential. For Z zmin , the Van der Waals forces are dominant. (b) For Z ≤ zmin , the wave function φ(Z) is such that φ(Z) = 0. Since φ(Z) is continuous, we have φ(zmin ) = 0. (c) For a turning point b, the WKB eigenfunction of energy E has the following form in the allowed region (E > V0 (Z)): φ(Z) =

C k(Z)

b

cos Z

π k(Z ) dZ − 4

where C is a normalization constant and where

,

25.5 Solutions

hk(Z) = ¯

261

2m(E − V0 (Z)) .

Imposing the condition φ(zmin ) = 0 yields

b

π k(Z ) dZ − = 4 zmin

1 n+ 2

3 k(Z ) dZ = n + π 4 zmin

π , i.e.

b

with n a positive integer. (d) If the WKB method were exact, the number of bound states would be

∞ k(Z ) 3 dZ − n = 1 + Int , π 4 zmin where Int denotes the integer part and where k(Z) is calculated for a zero energy E. As usual for the WKB method, the accuracy of this expression is good if is large. We can take in this case: ∞the number of bound states ¯ k(Z) = 2mα/Z 3 , which yields n π −1 zmin k(Z ) dZ with h 2mα 2 . n π¯ h zmin (e) The above formula yields n 1.36. We therefore expect a number of bound states close to 1, say between 0 and 2. (f ) The experimental result compares favorably with the WKB. prediction, but it is beyond the validity of the WKB approximation to give a correct expression for φ0 (Z). Section 25.2: Excitations on the Surface of Liquid Helium 25.2.1. The two terms of the dispersion relation are equal if q = gρ0 /A or, equivalently, for a wavelength A . gρ0

λ = 2π

Numerically, one obtains λ = 3 mm. Therefore, we observe capillary waves (ωq2 Aq 3 /ρ0 ) for λ 3 mm, and gravity waves (ωq2 gq) for λ 3 mm. For λ = 3 mm the corresponding energy is ¯hωq = 1.3 × 10−13 eV. 25.2.2. For an energy such that |E0 | 10−13 eV, we are therefore in the regime of capillary waves, with the wavelength: 2π = 2π λ= q The numerical value is λ = 33 ˚ A.

Ah2 ρ0 E02

1/3 .

262

25 Quantum Reﬂection of Atoms from a Surface

25.2.3. We have ˆ 2 − h ˆ 2 = h ˆ 2 = ∆h2 = h

h2q rq rq† =

q

h2q .

q

Converting this into an integral, we obtain qmax Lx Ly ¯h hq ¯ √ 2 √ ∆h2 = d q = q dq 2 4π 2ρ0 ωq Lx Ly 4π Aρ0 0 =

¯ h 6π

3 qmax hωmax ¯ , = Aρ0 6πA

which yields ∆h = 0.94 ˚ A. Section 25.3: Quantum Interaction Between H and Liquid He ˆ 25.3.1. Using the fact that Θ (z) = δ(z), we can write Θ(−z + h(r)) ˆ Θ(−z) + h(r)δ(z) since the δ function is even. Therefore, we obtain ˆ . V (R, Z) V0 (Z) + n d2 r U ( (R − r)2 + Z 2 ) h(r) In this expression, the second term describes the interaction with the “additional” or “missing” atoms on the surface as compared to the equilibrium position z = 0. ˆ 25.3.2. Replacing h(r) by its expansion we obtain V (R, Z) V0 (Z) + n d2 r U ( (R − r)2 + Z 2 ) hq (ˆ rq† e−iq·r + rˆq eiq·r ) . q

Considering the term rq† and setting r = r−R, we obtain in a straightforward manner

hq e−iq·R Vq (Z)ˆ rq† + h.c. , V (R, Z) = V0 (Z) + q

with Vq (Z) = n

d2 r e−iq·r U ( r 2 + Z 2 ) .

ˆ at = 25.3.3. The Hamiltonian is the sum of the “free” Hamiltonians H ˆ S , and the coupling term found above. One has V0 (Z) and H ˆS = ˆ at = H H Ek,σ a ˆ†k,σ a ˆk,σ ¯hωq rˆq† rˆq . k,σ

The coupling term becomes

q

P2 2M

+

25.5 Solutions

k,σ

k ,σ

q

263

hq a ˆ†k,σ a ˆk ,σ rˆq† k, φσ |e−iq.R Vq (Z)|k , φσ + h.c.

The matrix element is k, φσ |e−iq·R Vq (Z)|k , φσ = k|e−iq·R |k φσ |Vq (Z)|φσ = δk ,k+q φσ |Vq (Z)|φσ . ˆ We end up with the total hydrogen–ripplon Hamiltonian to ﬁrst order in h: ˆ = H Ek,σ a ˆ†k,σ a ˆk,σ + hωq rˆq† rq ¯ q

k,σ

+

q,k,σ,σ

hq a ˆ†k,σ a ˆk+q,σ rˆq†

φσ |Vq (Z)|φσ + h.c. .

In the (x, y) plane, the momentum is conserved owing to the translation invariance of the problem. This can be seen directly on the form of the coupling a ˆ†k,σ a ˆk+q,σ rˆq† , which annihilates a H atom with momentum h ¯ (k + q) term, and creates a H atom with momentum h ¯ k and a ripplon with momentum h ¯ q. Section 25.4: The Sticking Probability 25.4.1. We have by deﬁnition φ0 |Vq |φσ =

φ∗0 (Z) Vq (Z) φσ (Z) dZ .

Since |φσ is an asymptotically free state, it is normalized in a segment of −1/2 length Lz . Therefore its amplitude varies as Lz . Since |φ0 is a localized state which does not depend on Lz , we ﬁnd 1 φ0 |Vq |φσ ∝ √ . Lz The fact that this matrix element is proportional to kσ in the limit of small incident momenta is more subtle. The positions Z contributing to the matrix element are close to zero, since the bound state φ0 (Z) is localized in the vicinity of the He surface. Therefore only the values of φσ (Z) around Z = 0 are relevant for the calculation of the integral. For the Z −3 potential between the H atom and the He surface, one ﬁnds that the amplitude of φσ in this region is proportional to kσ , hence the result. Such a linear dependance can be recovered analytically by replacing the Z −3 potential by a square well, but is out √ of reach of the WKB approximation, which would predict a dependance in kσ for the amplitude around Z = 0 of φσ . The reason for this discrepancy is that the potential in −αZ −3 is too stiﬀ for the WKB to be valid for the calculation of φσ at distances larger than mα/¯h2 .

264

25 Quantum Reﬂection of Atoms from a Surface

25.4.2. We start with the initial state k⊥ , φσ . If the atom sticks to the surface, the ﬁnal state along the z axis is |φ0 . The sticking proceeds via the emission of a ripplon of momentum h ¯ q and a change of the transverse mohq. mentum h ¯ k⊥ → ¯hk⊥ − ¯ (a) The continuum of ﬁnal states is characterized by the vector q: |k⊥ , φσ → |k⊥ − q, φ0 ⊗ |q .

(b) Energy conservation gives Ei = Ef with:

Ei =

2 ) ¯ 2 (kσ2 + k⊥ h 2m

Ef = E0 +

¯ 2 (k⊥ − q)2 h +h ¯ ωq . 2m

We suppose that Ei is negligible compared to the bound state energy E0 . 2 /(2m) is also very small compared to Therefore ¯h2 (k⊥ · q)/m ∼ |E0 |¯ h2 k⊥ |E0 |, and we obtain: ¯ 2 q2 h +h ¯ ωq |E0 | . 2m This equation, in addition to the dispersion relation for ripplons, determines the modulus q0 of q: h ¯

A 3/2 h ¯ 2 q02 = |E0 | . q0 + ρ0 2m

(c) A variation δE of the ﬁnal state energy corresponds to a variation δq such that 3¯ h Aq0 h 2 q0 ¯ + δq = δE . m 2 ρ0 The number of states δ 2 n in a domain δ 2 q is:

δ2 n =

Lx Ly 2 Lx Ly δ q= q0 δq δθ . 2 4π 4π 2

After integrating over δθ, we obtain:

ρ(Ef ) =

mq0 Lx Ly . 2 π 2¯ h q0 + 3m¯h Aq0 /ρ0

25.6 Comments

265

25.4.3. The number of atoms which cross a plane of altitude Z in the direc¯ kσ dt/(2mLz ). The ﬂux tion Z < 0 during a time interval dt is vz dt/(2Lz ) = h is therefore: hkσ ¯ . Φσ = 2mLz 25.4.4. The sticking probability is the ratio of the probability per unit time, given by Fermi’s Golden Rule, and the incident ﬂux: P =

2π 2mLz |k⊥ , φσ |V |k⊥ − q, φ0 , q|2 ρ(Ef ) . h ¯ ¯hkσ

This reduces to P =

mkσ |M (q)|2 √ . 3Am + 2¯ h Aρ0 q0

√ 25.4.5. P varies as kσ ∝ E. At very small energies, the sticking probability goes to zero and the H atoms bounce elastically on the liquid He surface. 25.4.6. If the liquid helium bath is not at zero temperature, other processes can occur, in particular a sticking process accompanied by the stimulated emission of a ripplon. One must therefore take into account the number nq0 of thermal ripplons.

25.6 Comments The theory of the sticking of H atoms onto a surface of liquid He can be found in the [1] below. Thorough experimental studies of this process are presented in [2] and [3]. 1. D.S. Zimmerman and A.J. Berlinsky, Can. J. Phys. 61, 508 (1983). 2. J.J. Berkhout, O.J. Luiten, J.D. Setija, T.W. Hijmans, T. Mizusaki, and J.T.M. Walraven, Phys. Rev. Lett. 63, 1689 (1989). 3. J.M. Doyle, J.C. Sandberg, I.A. Yu, C.L. Cesar, D. Kleppner, and T.J. Greytak, Phys. Rev. Lett. 67, 603 (1991).

26 Laser Cooling and Trapping

By shining laser light onto an assembly of neutral atoms or ions, it is possible to cool and trap these particles. In this chapter we study a simple cooling mechanism, Doppler cooling, and we derive the corresponding equilibrium temperature. We then show that the cooled atoms can be conﬁned in the potential well created by a focused laser beam. We consider a “two state” atom, whose levels are denoted |g (ground state) and |e (excited state), with respective energies 0 and ¯hω0 . This atom interacts with a classical electromagnetic wave of frequency ωL /2π. For an atom located at r, the Hamiltonian is ˆ =h H ¯ ω0 |ee| − d · (E(r, t)|eg| + E ∗ (r, t)|ge|) ,

(26.1)

where d, which is assumed to be real, represents the matrix element of the atomic electric dipole operator between the states |g and |e (i.e. d = ˆ ˆ ∗ ). The quantity E + E ∗ represents the electric ﬁeld. We e|D|g = g|D|e set E(r, t) = E 0 (r) exp(−iωL t) . In all the chapter we assume that the detuning ∆ = ωL −ω0 is small compared with ωL and ω0 . We treat classically the motion r(t) of the atomic center of mass.

26.1 Optical Bloch Equations for an Atom at Rest 26.1.1. Write the evolution equations for the four components of the density operator of the atom ρgg , ρeg , ρge and ρee under the eﬀect of the Hamiltonian ˆ H. 26.1.2. We take into account the coupling of the atom with the empty modes of the radiation ﬁeld, which are in particular responsible for the spontaneous emission of the atom when it is in the excited state |e. We shall assume that this boils down to adding to the above evolution equations “relaxation” terms:

268

26 Laser Cooling and Trapping

d d ρee ρgg =− = −Γ ρee dt dt relax relax

d d Γ Γ ρeg ρge = − ρeg = − ρge , dt 2 dt 2 relax relax

where Γ −1 is the radiative lifetime of the excited state. Justify qualitatively these terms. 26.1.3. Check that for times much larger than Γ −1 , these equations have the following stationary solutions: s 2(s + 1) 2+s = 2(s + 1)

d · E(r, t)/¯ h 1 ∆ + iΓ/2 1 + s h 1 d · E ∗ (r, t)/¯ =− ∆ − iΓ/2 1 + s

ρee =

ρeg = −

ρgg

ρge

where we have set s=

h2 2 |d · E 0 (r)|2 /¯ . 2 2 ∆ + Γ /4

26.1.4. Interpret physically the steady state value of the quantity Γ ρee in terms of spontaneous emission rate.

26.2 The Radiation Pressure Force In this section, we limit ourselves to the case where the electromagnetic ﬁeld is a progressive plane wave: E 0 (r) = E 0 exp(ik · r) . By analogy with the classical situation, we can deﬁne the radiative force operator at point r as: ˆ (r) = −∇r H ˆ . F ˆ (r) assuming that the atom is at 26.2.1. Evaluate the expectation value of F rest in r and that its internal dynamics is in steady state. 26.2.2. Interpret the result physically in terms of momentum exchanges between the atom and the radiation ﬁeld. One can introduce the recoil velocity vrec = h ¯ k/m. 26.2.3. How does this force behave at high intensities? Give an order of magnitude of the possible acceleration for a sodium atom 23 Na, with a resonance wavelength λ = 0.589 × 10−6 m and a lifetime of the excited state of Γ −1 = 16 × 10−9 s (d = 2.1 × 10−29 C m). 26.2.4. We now consider an atom in uniform motion: r = r 0 + v 0 t (v0 c). Give the expression for the force acting on this atom.

26.3 Doppler Cooling

269

26.2.5. The action of the force on the atom will modify its velocity. Under what condition is it legitimate to treat this velocity as a constant quantity for the calculation of the radiation pressure force, as done above? Is this condition valid for sodium atoms?

26.3 Doppler Cooling The atom now moves in the ﬁeld of two progressive plane waves of opposite directions (+z and −z) and of same intensity (Fig. 26.1). We restrict ourselves to the motion along the direction of propagation of the two waves and we assume that for weak intensities (s 1) one can add independently the forces exerted by the two waves.

Fig. 26.1. Doppler cooling in one dimension

26.3.1. Show that for suﬃciently small velocities, the total force is linear in the velocity and can be cast in the form: f =−

mv . τ

26.3.2. What is the minimal (positive) value of τmin for a ﬁxed saturation parameter per wave s0 for an atom at rest? Calculate τmin for sodium atoms, assuming one ﬁxes s0 = 0.1. 26.3.3. This cooling mechanism is limited by the heating due to the random nature of spontaneous emission. To evaluate the evolution of the velocity distribution P (v, t) and ﬁnd its steady state value, we shall proceed in the following way: (a) Express P (v, t + dt) in terms of P (v, t). One will split the atoms into three classes: • the atoms having undergone no photon scattering event between t and t + dt, • the atoms having scattered a photon from the +z wave, • the atoms having scattered a photon from the −z wave. We choose dt short enough that the probability of the ﬁrst option is dominant, and such that multiple scattering events are negligible. We also assume that the velocities contributing signiﬁcantly to P (v, t) are

270

26 Laser Cooling and Trapping

small enough for the linearization of the force performed above to be valid. For simplicity we will assume that spontaneously emitted photons propagate only along the z axis, a spontaneous emission occurring with equal probabilities in the directions +z and −z. (b) Show that P (v, t) obeys the Fokker–Planck equation ∂ ∂2P ∂P = α (vP ) + β 2 ∂t ∂v ∂v and express of α and β in terms of the physical parameters of the problem. (c) Determine the steady state velocity distribution. Show that it corresponds to a Maxwell distribution and give the eﬀective temperature. (d) For which detuning is the eﬀective temperature minimal? What is this minimal temperature for sodium atoms?

26.4 The Dipole Force We now consider a stationary light wave (with a constant phase) E 0 (r) = E ∗0 (r) . ˆ (r) = 26.4.1. Evaluate the expectation value of the radiative force operator F ˆ −∇r H assuming that the atom is at rest in r and that its internal dynamics has reached its steady state. 26.4.2. Show that this force derives from a potential and evaluate the potential well depth that can be attained for sodium atoms with a laser beam of intensity P = 1 W, focused on a circular spot of radius 10 µm, and a wavelength λL = 0.650 µm.

26.5 Solutions Section 26.1: Optical Bloch Equations for an Atom at Rest 26.1.1. The evolution of the density operator ρˆ is given by: i¯ h

dˆ ρ ˆ ρˆ] = [H, dt

so that: dρee d · E(r) e−iωL t d · E ∗ (r) eiωL t =i ρge − i ρeg dt h ¯ ¯h dρeg d · E(r) e−iωL t = −iω0 ρeg + i (ρgg − ρee ) dt h ¯

26.5 Solutions

and dρee dρgg =− dt dt

dρge = dt

dρeg dt

271

∗ .

26.1.2. Assume that the atom-ﬁeld system is placed at time t = 0 in the state |ψ(0) = (α|g + β|e) ⊗ |0 , where |0 denotes the vacuum state of the electromagnetic ﬁeld and neglect in a ﬁrst step the action of the laser. At time t, the state of the system is derived from the Wigner–Weisskopf treatment of spontaneous emission: |ψ(t) = (α|g + βe−(iω0 +Γ/2)t) |e) ⊗ |0 + |g ⊗ |φ , where the state of the ﬁeld |φ is a superposition of one-photon states for the various modes of the electromagnetic ﬁeld. Consequently the evolution of the density matrix elements is ρee (t) = |β|2 e−Γ t , ρeg (t) = α∗ βe−(iω0 +Γ )t , or, in other words,

dρeg dρee Γ = −Γ ρee = − ρeg . dt relax dt relax 2 The two other relations originate from the conservation of the trace of the density operator (ρee + ρgg = 1) and from its hermitian character ρeg = ρ∗ge . We assume in the following that the evolution of the atomic density operator is obtained by adding the action of the laser ﬁeld and the spontaneous emission contribution. Since Γ varies like ω03 , this is valid as long as the shift of the atomic transition due to the laser irradiation remains small compared hω0 , which is satisﬁed for usual continuous laser with ω0 . This requires dE ¯ sources. 26.1.3. The evolution of the density operator components is given by d · E(r) e−iωL t d · E ∗ (r) eiωL t dρee = −Γ ρee + i ρge − i ρeg dt h ¯ ¯h

dρeg Γ d · E(r) e−iωL t = −iω0 − (ρgg − ρee ) . ρeg + i dt 2 h ¯ These equations are often called optical Bloch equations. At steady-state, ρee and ρgg tend to a constant value, while ρeg and ρge oscillate respectively as e−iωL t and eiωL t . This steady-state is reached after a characteristic time of the order of Γ −1 . ¿From the second equation we extract the steady-state value of ρeg as a function of ρgg − ρee = 1 − 2ρee : ρeg = i

h d · E(r) e−iωL t /¯ (1 − 2ρee ) . i∆ + Γ/2

We now insert this value in the evolution of ρee and we get:

272

26 Laser Cooling and Trapping

ρee =

s 2(1 + s)

with

s(r) =

h2 2 |d · E(r)|2 /¯ . ∆2 + Γ 2 /4

The three other values given in the text for ρgg , ρeg and ρge follow immediately. 26.1.4. The steady state value of ρee gives the average probability of ﬁnding the atom in the internal state |e. This value results from the competition between absorption processes, which tend to populate the level |e and stimulated+spontaneous emission processes, which depopulate |e to the beneﬁt of |g. The quantity Γ ρee represents the steady-state rate of spontaneous emission as the atom is irradiated by the laser wave. For a low saturation parameter s, this rate is proportional to the laser intensity |E(r)|2 . When the laser intensity increases, s gets much larger than 1 and the steady state value of ρee is close to 1/2. This means that the atom spends half of the time in level |e. In this case, the rate of spontaneous emission tends to Γ/2. Section 26.2: The Radiation Pressure Force 26.2.1. For a plane laser wave the force operator is given by: ˆ (r) = ik d · E 0 ei(k·r−ωL t) |eg| − e−i(k·r−ωL t) |ge| . F ˆ ) which gives: The expectation value in steady state is Tr(ˆ ρF f = F = ik d · E 0 ei(k·r−ωL t) ρge + c.c. s0 Γ =h ¯k 2 1 + s0 with s0 =

h2 2 |d · E 0 |2 /¯ . 2 2 ∆ + Γ /4

26.2.2. The interpretation of this result is as follows. The atom undergoes absorption processes, where it goes from the ground internal state to the excited internal state, and gains the momentum h ¯ k. From the excited state, it can return to the ground state by a stimulated or spontaneous emission process. In a stimulated emission the atom releases the momentum that it has gained during the absorption process, so that the net variation of momentum in a such a cycle is zero. In contrast, in a spontaneous emission process, the momentum change of the atom has a random direction and it averages to zero since the spontaneous emission process occurs with the same probability in two opposite directions. Therefore the net momentum gain for the atom in a cycle “absorption–spontaneous emission” is ¯hk corresponding to a velocity change vrec . Since these cycles occur with a rate (Γ/2)s0 /(1 + s0 ) (as found at the end of Sect. 26.1), we recover the expression for the radiation force found above.

26.5 Solutions

273

26.2.3. For a large laser intensity, the force saturates to the value ¯hkΓ/2. ¯ kΓ/(2m) = 9 × 105 m s−2 , which This corresponds to an acceleration amax = h is 100 000 times larger than the acceleration due to gravity. 26.2.4. In the rest frame of the atom, the laser ﬁeld still corresponds to a plane wave with a modiﬁed frequency ωL − k · v (ﬁrst order Doppler eﬀect). The change of momentum of the photon is negligible for non-relativistic atomic velocities. The previous result is then changed into: f =h ¯k

s(v) Γ 2 1 + s(v)

with

s(v) =

2 |d · E 0 |2 /¯ h2 . (∆ − k · v)2 + Γ 2 /4

26.2.5. The notion of force derived above is valid if the elementary velocity change in a single absorption or emission process (the recoil velocity vrec = h ¯ k/m) modiﬁes only weakly the value of f . This is the case when ¯ k 2 /m is very small compared the elementary change of Doppler shift kvrec = h with the width of the resonance: ¯ k2 h Γ . m This is the so called broad line condition. This condition is well satisﬁed for sodium atoms since ¯hk 2 /(mΓ ) = 5 × 10−3 in this case. Section 26.3: Doppler Cooling 26.3.1. The total force acting on the atom moving with velocity v is

|d · E 0 |2 /¯ |d · E 0 |2 /¯ h2 h2 f (v) = h ¯k Γ − , (∆ − kv)2 + Γ 2 /4 (∆ + kv)2 + Γ 2 /4 where we have used the fact that s 1. For low velocities (kv Γ ) we get at ﬁrst order in v f (v) = −

mv τ

with

τ=

m ∆2 + Γ 2 /4 . hk 2 s0 2(−∆)Γ ¯

This corresponds to a damping force if the detuning ∆ is negative. In this case the atom is cooled because of the Doppler eﬀect. This is the so-called Doppler cooling: A moving atom feels a stronger radiation pressure force from the counterpropagating wave than from the copropagating wave. For an atom at rest the two radiation pressure forces are equal and opposite: the net force is zero. 26.3.2. For a ﬁxed saturation parameter s0 , the cooling time is minimal for ∆ = −Γ/2, which leads to τmin =

m . 2¯ hk 2 s0

274

26 Laser Cooling and Trapping

Note that this time is always much longer than the lifetime of the excited state Γ −1 when the broad line condition is fullﬁlled. For sodium atoms this minimal cooling time is 16 µs for s0 = 0.1. 26.3.3. (a) The probability that an atom moving with velocity v scatters a photon from the ±z wave during the time dt is Γ s0 dP± (v) = 2

2∆kv 1± 2 ∆ + Γ 2 /4

dt .

Since we assume that the spontaneously emitted photons also propagate along z, half of the scattering events do not change the velocity of the atom: This is the case when the spontaneously emitted photon propagates along the same direction as the absorbed photon. For the other half of the events, the change of the atomic velocity is ±2vrec , corresponding to a spontaneously emitted photon propagating in the direction opposite to the absorbed photon. Consequently, the probability that the velocity of the atom does not change during the time dt is 1 − (dP+ (v) + dP− (v))/2, and the probability that the atomic velocity changes by ±2vrec is dP± (v)/2. Therefore one has:

P (v, t + dt) =

dP+ (v) + dP− (v) P (v, t) 2 dP+ (v − 2vrec ) P (v − 2vrec , t) + 2 dP− (v + 2vrec ) P (v + 2vrec , t) . + 2 1−

(b) Assuming that P (v) varies smoothly over the recoil velocity scale (which will be checked in the end), we can transform the ﬁnite diﬀerence equation found above into a diﬀerential equation: ∂ ∂2P ∂P = α (vP ) + β 2 , ∂t ∂v ∂v with m 2 β = Γ vrec s0 . τ The term proportional to α corresponds to the Doppler cooling. The term in β accounts for the heating due to the random nature of spontaneous emission processes. The coeﬃcient β is a diﬀusion constant in velocity space, proportional to the square of the elementary step of the random walk vrec , and to its rate Γ s0 . (c) The steady state for P (v) corresponds to the solution of: α=

26.5 Solutions

275

dP =0, dv whose solution (for α > 0, i.e. ∆ < 0) is a Maxwell distribution:

αv 2 P (v) = P0 exp − . 2β α v P (v) + β

The eﬀective temperature is therefore kB T =

h ∆2 + Γ 2 /4 ¯ mβ = . α 2 −∆

(d) The minimal temperature is obtained for ∆ = −Γ/2: ¯Γ h . 2 This is the Doppler cooling limit, which is independent of the laser intensity. Note that, when the broad line condition is fullﬁlled, the corresponding velocity scale v0 is such that: hΓ/m Γ/k . vrec v0 = ¯ kB Tmin =

The two hypotheses at the basis of our calculation are therefore valid: (i) P (v) varies smootly over the scale vrec and (ii) the relevant velocities are small enough for the linearization of the scattering rates to be possible. For sodium atoms, the minimal temperature is Tmin = 240 µK, corresponding to v0 = 40 cm s−1 . Section 26.4: The Dipole Force 26.4.1. For a real amplitude E 0 (r) of the electric ﬁeld of the light wave ˆ (r) is: (standing wave), the force operator F ⎞ ⎛

ˆ (r) = ⎝ di ∇E0i (r)⎠ e−iωL t |eg| + eiωL t |ge| . F i=x,y,z

Assuming that the internal dynamics of the atom has reached its steady-state ˆ: value, we get for the expectation value of F f (r) = F = −∇(d · E 0 (r)) =−

∆ d · E 0 (r) 2 1 + s(r) ∆ + Γ 2 /4

¯ ∆ ∇s(r) h 2 1 + s(r)

26.4.2. This force is called the dipole force. It derives from the dipole potential U (r):

276

26 Laser Cooling and Trapping

f (r) = −∇U (r)

with

U (r) =

¯∆ h log(1 + s(r)) . 2

For a laser ﬁeld with intensity P = 1 W, focused on a spot with radius r = 10 µm, the electric ﬁeld at the center is 2P E0 = = 1.6 × 106 V/m . π0 cr2 We suppose here that the circular spot is uniformly illuminated. A more accurate treatment should take into account the transverse Gaussian proﬁle of the laser beam, but this would not signiﬁcantly change the following reh = 3.1 × 1011 s−1 and the detuning ∆ is sults. This value for E0 leads to dE0 /¯ 14 −1 equal to 3×10 s . The potential depth is then found to be equal to 2.4 mK, 10 times larger than the Doppler cooling limit. Due to the large detuning, the photon scattering rate is quite small: 70 photons/s.

26.6 Comments The radiation pressure force has been used in particular for atomic beam deceleration (J.V. Prodan, W.D. Phillips, and H. Metcalf, Phys. Rev. Lett. 49, 1149 (1982)). The Doppler cooling was proposed by T.W. H¨ ansch and A. Schawlow (Opt. Commun. 13, 68 (1975)). A related cooling scheme for trapped ions was proposed the same year by D. Wineland and H. Dehmelt (Bull. Am. Phys. Soc. 20, 637 (1975)). The ﬁrst observation of 3D laser cooling of neutral atoms was reported by S. Chu, L. Hollberg, J.E. Bjorkholm, A. Cable, and A. Ashkin, Phys. Rev. Lett. 55, 48 (1985), and the same group reported one year later the observation of atoms trapped at the focal point of a laser beam using the dipole force (Phys. Rev. Lett. 57, 314 (1986)). It was subsequently discovered experimentally in the group of W.D. Phillips that the temperature of laser cooled atoms could be much lower than ¯ Γ/2 derived in this problem. This clear violation the Doppler limit kB T = h of Murphy’s law (an experiment working 10 times better than predicted!) was explained independently in terms of Sisyphus cooling by the groups of C. Cohen-Tannoudji and S. Chu (for a review, see e.g. C. Cohen-Tannoudji and W.D. Phillips, Physics Today, October 1990, p.33). The Physics Nobel Prize was awarded in 1997 to S. Chu, C. CohenTannoudji and W.D. Phillips for their work on the trapping and cooling of atoms with laser light.

27 Bloch Oscillations

The possibility to study accurately the quantum motion of atoms in standing light ﬁelds has been used recently in order to test several predictions relating to wave propagation in a periodic potential. We present in this chapter some of these observations related to the phenomenon of Bloch oscillations.

27.1 Unitary Transformation on a Quantum System Consider a system in the state |ψ(t) which evolves under the eﬀect of a ˆ ˆ Hamiltonian H(t). Consider a unitary operator D(t). Show that the evolution of the transformed vector ˜ ˆ |ψ(t) = D(t)|ψ(t) is given by a Schr¨ odinger equation with Hamiltonian ˆ dD(t) ˜ ˆ † (t) . ˆ ˆ H(t) ˆ D ˆ † (t) + i¯ D H(t) = D(t) h dt

27.2 Band Structure in a Periodic Potential The mechanical action of a standing light wave onto an atom can be described by a potential (see e.g. Chap. 26). If the detuning between the light frequency and the atom resonance frequency ωA is large compared to the electric dipole coupling of the atom with the wave, this potential is proportional to the light intensity. Consequently, the one-dimensional motion of an atom of mass m moving in a standing laser wave can be written ˆ2 ˆ , ˆ = P + U0 sin2 (k0 X) H 2m

278

27 Bloch Oscillations

ˆ and Pˆ are the atomic position and momentum operators and where where X we neglect any spontaneous emission process. We shall assume that k0 ωA /c ¯ 2 k02 /(2m). and we introduce the “recoil energy” ER = h ˆ recall brieﬂy why 27.2.1. (a) Given the periodicity of the Hamiltonian H, the eigenstates of this Hamiltonian can be cast in the form (Bloch theorem): |ψ = eiqX |uq , ˆ

where the real number q (Bloch index) is in the interval (−k0 , k0 ) and where |uq is periodic in space with period λ0 /2. (b) Write the eigenvalue equation to be satisﬁed by |uq . Discuss the corresponding spectrum (i) for a given value of q, (ii) when q varies between −k0 and k0 . ˆ are denoted |n, q, with energies En (q). In the following, the eigenstates of H They are normalized on a spatial period of extension λ0 /2 = π/k0 . 27.2.2. Give the energy levels in terms of the indices n and q in the case U0 = 0. 27.2.3. Treat the eﬀect of the potential U0 in ﬁrst order perturbation theory, for the lowest band n = 0 (one should separate the cases q = ±k0 and q “far from” ±k0 ). Give the width of the gap which appears between the bands n = 0 and n = 1 owing to the presence of the perturbation. 27.2.4. Under what condition on U0 is this perturbative approach reliable? 27.2.5. How do the widths of the other gaps vary with U0 in this perturbative limit?

27.3 The Phenomenon of Bloch Oscillations We suppose now that we prepare in the potential U0 sin2 (k0 x) a wave packet in the n = 0 band with a sharp distribution in q, and that we apply to the atom a constant extra force F = ma. We recall the adiabatic theorem: suppose that a system is prepared at time (0) ˆ ˆ If the Hamiltonian H(t) 0 in the eigenstate |φn of the Hamiltonian H(0). evolves slowly with time, the system will remain with a large probability in (t) (t) (t) ¯ φm |φ˙ n the eigenstate |φn . The validity condition for this theorem is h (t) (t) d |Em (t) − En (t)| for any m = n. We use the notation |φ˙ n = dt |φn . 27.3.1. Preparation of the Initial State. Initially U0 = 0, a = 0 and the atomic momentum distribution has a zero average and a dispersion small

27.3 The Phenomenon of Bloch Oscillations

279

compared to ¯hk. We will approximate this state by the eigenstate of momentum |p = 0. One “slowly” switches on the potential U0 (t) sin2 (k0 x), with U0 (t) ≤ ER . (a) Using the symmetries of the problem, show that the Bloch index q is a constant of the motion. (b) Write the expression of the eigenstate of H(t) of indices n = 0, q = 0 to ﬁrst order in U0 . (c) Evaluate the validity of the adiabatic approximation in terms of U˙ 0 , ER , ¯h. (d) One switches on linearly the potential U0 until it reaches the value ER . What is the condition on the time τ of the operation in order for the process to remain adiabatic? Calculate the minimal value of τ for cesium atoms (m = 2.2 × 10−25 kg, λ0 = 0.85 µm.) 27.3.2. Devising a Constant Force. Once U0 (t) has reached the maximal value U0 (time t = 0), one achieves a sweep of the phases φ+ (t) and φ− (t) of the two traveling waves forming the standing wave. The potential seen by the atom is then U0 sin2 (k0 x − (φ+ (t) − φ− (t))/2) and one chooses φ+ (t) − φ− (t) = k0 at2 . (a) Show that there exists a reference frame where the wave is stationnary, and give its acceleration. (b) In order to study the quantum motion of the atoms in the accelerated reference frame, we consider the unitary transformation generated by ˆ h) exp(ima2 t3 /(3¯ ˆ h) exp(−imatX/¯ D(t) = exp(iat2 Pˆ /2¯ h)) . ˆ and Pˆ transform? Write How do the position and momentum operators X the resulting form of the Hamiltonian in this unitary transformation ˆ2 ˜ ˆ = P + U0 sin2 (k0 X) ˆ + maX ˆ . H 2m

27.3.3. Bloch Oscillations We consider the evolution of the initial state n = 0, q = 0 under the eﬀect of ˜ ˆ the Hamiltonian H. (a) Check that the state vector remains of the Bloch form, i.e. |ψ(t) = eiq(t)X |u(t) , ˆ

where |u(t) is periodic in space and q(t) = −mat/¯h. (b) What does the adiabatic approximation correspond to for the evolution of |u(t)? We shall assume this approximation to be valid in the following.

280

27 Bloch Oscillations

(c) Show that, up to a phase factor, |ψ(t) is a periodic function of time, and give the corresponding value of the period. (d) The velocity distribution of the atoms as a function of time is given in Fig. 27.1. The time interval between two curves is 1 ms and a = −0.85 ms−2 . Comment on this ﬁgure, which has been obtained with cesium atoms.

Fig. 27.1. Atomic momentum distribution of the atoms (measured in the accelerated reference frame) for U0 = 1.4 ER . The lower curve corresponds to the end of the preparation phase (t = 0) and the successive curves, from bottom to top, are separated by time intervals of one millisecond. For clarity, we put a diﬀerent vertical oﬀset for each curve

27.4 Solutions

281

27.4 Solutions Section 27.1: Unitary Transformation on a Quantum System ˜ gives The time derivative of |ψ ˙ = i¯ ˜˙ = i¯ ˆ˙ D ˆ† + D ˆ˙ ˆ ψ ˆ ˆH ˆD ˆ † D|ψ + D| h D h D|ψ , i¯ h|ψ hence the results of the lemma. Section 27.2: Band Structure in a Periodic Potential 27.2.1. Bloch theorem (a) The atom moves in a spatially periodic potential, with period λ0 /2 = π/k0 . Therefore the Hamiltonian commutes with the translation operator h)) and one can look for a common basis set of Tˆ(λ0 /2) = exp(iλ0 Pˆ /(2¯ these two operators. Eigenvalues of Tˆ(λ0 /2) have a modulus equal to 1, since Tˆ(λ0 /2) is unitary. They can be written eiqλ0 /2 where q is in the interval ˆ and Tˆ(λ0 /2) is then such that (−k0 , k0 ). A corresponding eigenvector of H Tˆ(λ0 /2)|ψ = eiqλ0 /2 |ψ or in other words ψ(x + λ0 /2) = eiqλ0 /2 ψ(x) . This amounts to saying that the function uq (x) = e−iqx ψ(x) is periodic in space with period λ0 /2, hence the result. (b) The equation satisﬁed by uq is: −

¯2 h 2m

d + iq dx

2 uq + U0 sin2 (kx) uq = E uq .

For a ﬁxed value of q, we look for periodic solutions of this equation. The boundary conditions uq (λ0 /2) = uq (0) and uq (λ0 /2) = uq (0) lead, for each q, to a discrete set of allowed values for E, which we denote En (q). The ˆ and Tˆ(λ0 /2) is denoted |ψ = |n, q. Now corresponding eigenvector of H when q varies in the interval (−k0 , k0 ), the energy En (q) varies continuously in an interval (Enmin , Enmax ). The precise values of Enmin and Enmax depend on the value of U0 . The spectrum En (q) is then constituted by a series of allowed energy bands, separated by gaps corresponding to forbidden values of energy. The interval (−k0 , k0 ) is called the ﬁrst Brillouin zone. ˆ is simply ¯h2 k 2 /(2m) corresponding 27.2.2. For U0 = 0, the spectrum of H ikx (free particle). Each k can be written: k = q + 2nk0 to the eigenstates e where n is an integer, and the spectrum En (q) then consists of folded portions of parabola (see Fig. 27.2a). There are no forbidden gaps in this case, and the min = Enmax ). various energy bands touch each other (En+1

282

27 Bloch Oscillations

Fig. 27.2. Structure of the energy levels En (q) (a) for U0 = 0 and (b) U0 = ER

ˆ has no degeneracy, 27.2.3. When q is far enough from ±k0 , the spectrum of H and the shift of the energy level of the lowest band n = 0 can be obtained using simply: ˆ ∆E0 (q) = 0, q|U0 sin2 (k0 X)|0, q π/k0 U0 k0 . e−iqx U0 sin2 (k0 x) eiqx dx = = π 0 2 When q is equal to ±k0 , the bands n = 0 and n = 1 coincide and one should diagonalize the restriction of U (x) to this two-dimensional subspace. One gets U0 ˆ ˆ k0 = 1, k0 |U0 sin2 (k0 X)|1, k0 = , 0, k0 |U0 sin2 (k0 X)|0, 2 U0 ˆ ˆ 0, k0 |U0 sin2 (k0 X)|1, k0 = 1, k0 |U0 sin2 (k0 X)|0, k0 = − . 4 The diagonalization of the matrix

U0 2 −1 4 −1 2 gives the two eigenvalues 3U0 /4 and U0 /4, which means that the two bands n = 0 and n = 1 do not touch each other anymore, but that they are separated by a gap U0 /2 (see Fig. 27.2b for U0 = ER ). 27.2.4. This perturbative approach is valid if one can neglect the coupling to all other bands. Since the characteristic energy splitting between the band n = 1 and the band n = 2 is 4 ER , the validity criterion is U0 4 ER . 27.2.5. The other gaps open either at k = 0 or k = ±k0 . They result from the coupling of eink0 x and e−ink0 x under the inﬂuence of U0 sin2 (k0 x). This coupling gives a non-zero result when taken at order n. Therefore the other gaps scale as U0n and they are much smaller that the lowest one.

27.4 Solutions

283

Section 27.3: The phenomenon of Bloch Oscillations 27.3.1. Preparation of the Initial State (a) Suppose that the initial state has a well deﬁned Bloch index q, which means that Tˆ(λ0 /2)|ψ(0) = eiqλ0 /2 |ψ(0) . ˆ At any time t, the Hamiltonian H(t) is spatially periodic and commute with ˆ (t) also the translation operator Tˆ(λ0 /2). Therefore the evolution operator U commutes with Tˆ(λ0 /2). Consequently: ˆ (t)|ψ(0) = U ˆ (t)Tˆ(λ0 /2)|ψ(0) , Tˆ(λ0 /2)|ψ(t) = Tˆ(λ0 /2)U = eiqλ0 /2 |ψ(t) , which means that q is a constant of motion. (b) At zeroth order in U0 , the eigenstates of H corresponding to the Bloch index q = 0 are the plane waves |k = 0 (energy 0), |k = ±2k0 (energy 4ER ), . . . At ﬁrst order in U0 , in order to determine |n = 0, q = 0, we have to take into account the coupling of |k = 0 with |k = ±2k0 , which gives |n = 0, q = 0 = |k = 0 +

k = 2k0 |U0 sin2 k0 x|k = 0 |k = 2k0 . 4ER =±

The calculation of the matrix elements is straightforward and it leads to x|n = 0, q = 0 ∝ 1 +

U0 (t) cos(2k0 x) . 8ER

(c) The system will adiabatically follow the level |n = 0, q = 0 as the potential U0 is raised, provided for any n

d|n = 0, q = 0 hn , q = 0| ¯ En (0) − E0 (0) . dt Using the value found above for |n = 0, q = 0 and taking n = ±1, we derive the validity criterion for the adiabatic approximation in this particular case: 2 . hU˙ 0 64 ER ¯

(d) For a linear variation of U0 such that U0 = ER t/τ , this validity condition is τ ¯ h/(64 ER ) , which corresponds to τ 10 µs for cesium atoms.

284

27 Bloch Oscillations

27.3.2. Devising a Constant Force (a) Consider a point with coordinate x in the lab frame. In the frame with acceleration a and zero initial velocity, the coordinate of this point is x = x − at2 /2. In this frame, the laser intensity varies as sin2 (kx ), corresponding to a “true” standing wave. ˆ f (Pˆ )] = i¯ ˆ = (b) Using the standard relations [X, hf (Pˆ ) and [Pˆ , g(X)] ˆ one gets −i¯ hg (X), 2 ˆ Pˆ D ˆ † = Pˆ + mat . ˆX ˆD ˆ† = X ˆ + at D D 2 ˆH ˆD ˆ † is The transformed Hamiltonian D

2 ˆ , ˆH ˆD ˆ † = 1 Pˆ + mat + U0 sin2 (k0 X) D 2m ˜ ˆ can be written and the extra term appearing in H 2 2 ˆ dD ˆ † = −atPˆ + maX ˆ − ma t . D dt 2 Summing the two contributions, we obtain

i¯ h

ˆ2 ˜ ˆ = P + U0 sin2 (k0 X) ˆ + maX ˆ . H 2m This Hamiltonian describes the motion of a particle of mass m in a periodic potential, superimposed with a constant force – ma. 27.3.3. Bloch Oscillations ˜ˆ ˙ = H|ψ. (a) The evolution of the state vector is i¯ h|ψ We now put |ψ(t) = ˆ h)|u(t) and we look for the evolution of |u(t). We obtain after exp(−imatX/¯ a straightforward calculation h2 ¯ ∂u(x, t) =− i¯ h ∂t 2m

imat ∂ − ∂x h ¯

2 u(x, t) + U0 sin2 (k0 x) u(x, t) .

Using the structure of this equation, and using the initial spatial periodicity of u(x, 0), one deduces that u(x, t) is also spatially periodic with the same period λ0 /2. (b) The adiabatic hypothesis for |u(t) amounts to assume that this vector, which is equal to |un=0,q=0 at t = 0, remains equal to |u0,q(t) at any time. The atom stays in the band n = 0. hk0 /(ma) during which q(t) is changed into (c) Consider the duration TB = 2¯ q(t) − 2k0 . Since 2k0 is the width of the Brillouin zone, we have |un,q−2k0 ≡ |un,q . Consequently, when the adiabatic approximation is valid, the state |ψ(t + TB ) coincides (within a phase factor) with the state |ψ(t). Since this phase factor does not enter in the calculation of physical quantities such as

27.5 Comments

285

position or momentum distributions, we expect that the evolution of these quantities with time will be periodic with the period TB . (d) We ﬁrst note that the initial distribution is such that the average momentum is zero, and that the momentum dispersion is small compared with hk0 , as assumed in this problem. Concerning the time evolution, we see in¯ deed that the atomic momentum distribution is periodic in time, with a period TB 8 ms, which coincides with the predicted value 2¯hk0 /(ma). Finally we note that the average momentum increases quasi-linearly with time during the ﬁrst 4 ms, from 0 to ¯hk0 . At this time corresponding to TB /2, a “reﬂexion” occurs and the momentum is changed into −¯ hk0 . During the second half of the Bloch period (from 4 ms to 8 ms) the momentum again increases linearly with time from −¯ hk0 to 0. At the time TB /2, the particle is at the edge of the Brillouin zone (q = ±k0 ). This is the place where the adiabatic approximation is the most fragile since the band n = 1 is then very close to the band n = 0 (gap U0 ). One can check that the validity criterion for the adiabatic approximation at this place is maER k0 U02 , which is well fullﬁlled in the experiment. The reﬂection occurring at t = TB /2 can be viewed as a Bragg reﬂection of the atom with momentum h ¯ k0 on the periodic grating U0 sin2 (kx).

27.5 Comments This paradoxical situation, where a constant force ma leads to an oscillation of the particle instead of a constant acceleration, is called the Bloch oscillation phenomenon. It shows that an ideal crystal cannot be a good conductor: when one applies a potential diﬀerence at the edge of the crystal, the electrons of the conduction band feel a constant force in addition to the periodic potential created by the crystal and they should oscillate instead of being accelerated towards the positive edge of the crystal. The conduction phenomenon results from the defects present in real metals. The experimental data have been extracted from M. Ben Dahan, E. Peik, J. Reichel, Y. Castin, and C. Salomon, Phys. Rev. Lett. 76, 4508 (1996) and from E. Peik, M. Ben Dahan, I. Bouchoule, Y. Castin, and C. Salomon, Phys. Rev. A 55, 2989 (1997). A review of atom optics experiments performed with standing light waves is given in M. Raizen, C. Salomon, and Q. Niu, Physics Today, July 1997, p. 30.

1 Neutrino Oscillations

In β decay or, more generally, in Weak interactions, the electron is always associated with a neutral particle, the neutrino νe . There exists in nature another particle, the µ lepton, or muon, whose physical properties seem completely analogous to those of the electron, except for its mass mµ 200 me . The muon has the same Weak interactions as the electron, but it is associated to a diﬀerent neutrino, the νµ . A neutrino beam produced in an accelerator can interact with a neutron (n) in a nucleus and give rise to the reactions νe + n → p + e

and

νµ + n → p + µ ,

(1.1)

whereas the reactions νe + n → p + µ or νµ + n → p + e are never observed. The reactions (1.1) are used in practice in order to detect neutrinos. Similarly, a π − meson can decay via the modes π − → µ + ν¯µ

(dominant mode) and π − → e + ν¯e ,

(1.2)

whereas π − → µ + ν¯e or π − → e + ν¯µ are never observed. This is how one can produce neutrinos abundantly (it is easy to produce π mesons). In (1.2) we have introduced the antiparticles ν¯µ et ν¯e . There is a (quasi) strict symmetry between particles and their antiparticles, so that, in the same way as the electron is associated with the neutrino νe , the antielectron, or positron, e+ is associated with the antineutrino ν¯e . One observes the “charge-conjugate” reactions of (1.1) and (1.2) ν¯e + p → n + e+ ,

ν¯µ + p → n + µ+

and π + → µ+ + νµ .

(1.3)

In all what follows, what we will say about neutrinos holds symmetrically for antineutrinos. In 1975, a third lepton, the τ , was discovered. It is much more massive, mµ 3500 me , it is associated with its own neutrino ντ , and it obeys the same physical laws as the two lighter leptons, except for mass eﬀects. Since the

18

1 Neutrino Oscillations

1990’s, the experimental measurements at the LEP colliding ring in CERN have shown that these three neutrinos νe , νµ , ντ (and their antiparticles) are the only ones of their kinds (at least for masses less that 100 GeV/c2 ). For a long time, physicists believed that neutrinos were zero-mass particles, as is the photon. In any case, their masses (multiplied by c2 ) are considerably smaller than the energies involved in experiments where they are observed. Therefore, many experimental limits on these masses are consistent with zero. However, both theoretical and cosmological arguments suggested that this might not be the case. The proof that neutrino masses are not all zero is a great discovery of the last ten years. In the present study, we show how the mass diﬀerences of neutrinos can be measured by a quantum oscillation eﬀect. The idea is that the “ﬂavor” neutrinos νe , νµ and ντ , which are produced or detected experimentally are not eigenstates of the mass, but rather linear combinations of mass eigenstates ν1 , ν2 , ν3 , with masses m1 , m2 , m3 . The neutrinos observed on earth have various origins. They can be produced in accelerators, in nuclear reactors, and also in the atmosphere by cosmic rays, or in thermonuclear reaction inside stars, in particular the core of the sun, and in supernovae explosions.

1.1 Mechanism of the Oscillations; Reactor Neutrinos In this ﬁrst part, we consider oscillations between two types of neutrinos, the νe and the νµ . This simple case will allow us to understand the underlying physics of the general case. We will analyze the data obtained with nuclear reactors. The average energy of the (anti-)neutrinos produced in reactors is E = 4 MeV, with a dispersion of the same order. In all what follows, we will assume that if m is the neutrino mass and p and E its momentum and energy, the mass is so small that the energy of a neutrino of mass m and momentum p is E=

p2 c2 + m2 c4 pc +

m2 c4 , 2pc

(1.4)

and that the neutrino propagates to very good approximation at the velocity of light c. ˆ be the Hamiltonian of a free neutrino of momentum p, which we Let H ˆ assume to be well deﬁned. We note |ν1 and |ν2 the two eigenstates of H: ˆ j = Ej |νj , H|ν

Ej = pc +

m2j c4 , 2pc

j = 1, 2 .

m1 and m2 are the respective masses of the states |ν1 and |ν2 , and we assume m1 = m2 .

1.1 Mechanism of the Oscillations; Reactor Neutrinos

19

The oscillations of freely propagating neutrinos come from the following quantum eﬀect. If the physical states of the neutrinos which are produced (reactions (1.2)) or detected (reactions (1.1)) are not |ν1 and |ν2 , but linear combinations of these: |νe = |ν1 cos θ + |ν2 sin θ ,

|νµ = −|ν1 sin θ + |ν2 cos θ

(1.5)

where θ is a mixing angle to be determined, these linear combination of energy eigenstates oscillate in time and this leads to measurable eﬀects. 1.1.1. At time t = 0, one produces a neutrino of momentum p in the state |νe . Calculate the state |ν(t) at time t in terms of |ν1 and |ν2 . 1.1.2. What is the probability Pe for this neutrino to be detected in the state |νe at time t? The result will be expressed in terms of the mixing angle θ and of the oscillation length L L=

4π¯ hp , |∆m2 | c2

∆m2 = m21 − m22 .

(1.6)

1.1.3. Calculate the oscillation length L for an energy E pc = 4 MeV and a mass diﬀerence ∆m2 c4 = 10−4 eV2 . 1.1.4. One measures the neutrino ﬂuxes with a detector located at a distance from the production area. Express the probability Pe as a function of the distance = ct. 1.1.5. The mass of the muon satisﬁes mµ c2 = 106 MeV. Conclude that in such an experiment one cannot detect muon neutrinos νµ with the reaction (1.1). We recall that mp c2 = 938.27 MeV and mn c2 = 939.57 MeV. 1.1.6. The detectors measure neutrino ﬂuxes with an accuracy of ∼ 10%. (a) Assuming ∆m2 c4 = 10−4 eV2 , determine the minimal distance min where to put a detector in order to detect an oscillation eﬀect. For this calculation, assume the mixing in (1.5) is maximum, i.e. θ = π/4. (b) How does min change if the mixing is not maximum? 1.1.7. Several experiments on neutrinos produced by nuclear energy plants have been performed in Chooz and in Bugey in France. The most recent data comes from the KamLAND collaboration, in Japan. The results are given on Fig. 1.1. (a) Explain the results of Fig. 1.1, except that of KamLAND. (b) The KamLAND experiment, which was performed in 2002, consisted in measuring the neutrinos coming from all the (numerous) reactors in Japan and neighboring countries, which amounts to taking an average distance of = 180 km. Putting together that data and the results of numerous experiments performed on solar neutrinos, the physicists of Kamland come to the following results:

20

1 Neutrino Oscillations

Ndetected N expected

1,2 1,0 0,8 0,6

ILL Bugey Rovno Goesgen Chooz KamLAND

0,4 0,2 0 10

102

distance (meters) 103

104

105

Fig. 1.1. Ratio between the numbers of observed electron neutrinos and those expected in the absence of oscillations as a function of the distance to the reactor

|∆m2 | c4 = 7.1 (± 0.4) × 10−5 eV2 ,

tan2 θ = 0.45 (± 0.02) . (1.7)

Show that these values are consistent with the result Pe = 0.61 (± 0.10) of Fig. 1.1.

1.2 Oscillations of Three Species; Atmospheric Neutrinos We now consider the general formalism with three neutrino species. We denote |να , α = e, µ, τ the “ﬂavor” neutrinos and |νi , i = 1, 2, 3 the mass eigenstates. These two bases are related to one another by the Maki-Nagawakaˆ, Sakata (MNS) matrix U ⎞ ⎛ Ue1 Ue2 Ue3 3 ⎟ ˆ =⎜ Uαi |νi , U |να = (1.8) ⎝ Uµ1 Uµ2 Uµ3 ⎠ i=1 Uτ 1 Uτ 2 Uτ 3 ∗ This matrix is unitary ( i Uβi Uαi = δαβ ) and it can be written as: ⎞ ⎞⎛ ⎞⎛ ⎛ 0 s13 e−iδ c13 c12 s12 0 1 0 0 ˆ = ⎝ 0 c23 s23 ⎠ ⎝ ⎠ ⎝ −s12 c12 0 ⎠ 0 1 0 U iδ 0 −s23 c23 0 0 1 −s13 e 0 c13 where cij = cos θij and sij = sin θij . The complete experimental solution of the problem would consist in measuring the three mixing angles θ12 , θ23 , θ13 , the phase δ, and the three masses m1 , m2 , m3 . We consider situations such that (1.4) is still valid. 1.2.1. At time t = 0 a neutrino is produced with momentum p in the state |ν(0) = |να . Express, in terms of the matrix elements Uαi , its state at a

1.2 Oscillations of Three Species; Atmospheric Neutrinos

21

later time t. Write the probability Pα→β (t) to observe a neutrino of ﬂavor β at time t. 1.2.2. We deﬁne the oscillation lengths at an energy E pc by: Lij =

4π¯ hp , |∆m2ij |c2

∆m2ij = m2i − m2j .

(1.9)

Notice that there are only two independent oscillation lengths since ∆m212 + ∆m223 + ∆m231 = 0. For neutrinos of energy E = 4 GeV, calculate the oscillation lengths L12 and L23 . We will choose for |∆m212 | the result given in (1.7), and we will choose |∆m223 | c4 = 2.5 × 10−3 eV2 , a value which will be justiﬁed later on. 1.2.3. The neutrino counters have an accuracy of the order of 10% and the energy is E = 4 GeV. Above which distances 12 and 23 of the production point of the neutrinos can one hope to detect oscillations coming from the superpositions 1 ↔ 2 and 2 ↔ 3? 1.2.4. The Super-Kamiokande experiment, performed in 1998, consists in detecting “atmospheric” neutrinos. Such neutrinos are produced in the collision of high energy cosmic rays with nuclei in the atmosphere at high altitudes. In a series of reactions, π ± mesons are produced abundantly, and they decay through the chain: π − → µ− + ν¯µ

followed by

µ− → e− + ν¯e + νµ ,

(1.10)

and an analogous chain for π + mesons. The neutrino ﬂuxes are detected in an underground detector by the reactions (1.1) and (1.3). To simplify things, we assume that all muons decay before reaching the surface of the Earth. Deduce that, in the absence of neutrino oscillations, the expected ratio between electron and muon neutrinos Rµ/e =

νµ ) N (νµ ) + N (¯ N (νe ) + N (¯ νe )

would be equal to 2. 1.2.5. The corrections to the ratio Rµ/e due to the fact that part of the muons reach the ground can be calculated accurately. Once this correction is made, one ﬁnds, by comparing the measured and calculated values for Rµ/e (Rµ/e )measured = 0.64 (± 0.05) . (Rµ/e )calculated In order to explain this relative decrease of the number of νµ ’s, one can think of oscillations of the types νµ νe and νµ ντ . The Super-Kamiokande experiment consists in varying the time of ﬂight of the neutrinos by measuring selectively the direction where they come from, as indicated on Fig. 1.2. The

22

1 Neutrino Oscillations

Electron neutrinos 120 cosmic ray

zenithal angle α

80 40

neutrino

cos α

Detector 0

Earth

-1

-0,5

0

0,5

1

Muon neutrinos

300

200 atmosphere 100 cos α 0

-1

-0,5

0

0,5

1

Fig. 1.2. Left: production of atmospheric neutrinos in collisions of cosmic rays with terrestrial atmospheric nuclei. The underground detector measures the ﬂux of electron and muon neutrinos as a function of the zenithal angle α. Right: number of atmospheric neutrinos detected in the Super-Kamiokande experiment as a function of the zenithal angle (this picture is drawn after K. Tanyaka, XXII Physics in Collisions Conference, Stanford 2002)

neutrinos coming from above (cos α ∼ 1) have traveled a distance equal to the atmospheric height plus the depth of the detector, while those coming from the bottom (cos α ∼ −1) have crossed the diameter of the Earth (13 400 km). Given the weakness of the interaction of neutrinos with matter, one can consider that the neutrinos propagate freely on a measurable distance between a few tens of km and 13 400 km. The neutrino energies are typically 4 GeV in this experiment. Can one observe a νe νµ oscillation of the type studied in the ﬁrst part? 1.2.6. The angular distributions of the νe and the νµ are represented on Fig. 1.2, together with the distributions one would observe in the absence of oscillations. Explain why this data is compatible with the fact that one observes a νµ ντ oscillation, no νe ντ oscillation, and no νe νµ oscillation.

1.3 Solutions

23

1.2.7. In view of the above results, we assume that there is only a twoneutrino oscillation phenomenon: νµ ντ in such an observation. We therefore use the same formalism as in the ﬁrst part, except that we change the names of particles. By comparing the muon neutrino ﬂux coming from above and from below, give an estimate of the mixing angle θ23 . In order to take into account the large energy dispersion of cosmic rays, and therefore of atmospheric neutrinos, we replace the oscillating factor sin2 (π/L23 ) by its mean value 1/2 if L23 . The complete results published by the Super-Kamiokande experiment are |∆m223 | c4 = 2.5 × 10−3 eV2 ,

θ23 = π/4 ,

θ13 = 0 .

Do they agree with the above considerations?

1.3 Solutions Section 1.1: Mechanism of the Oscillations: Reactor Neutrinos 1.1.1. Initially, the neutrino state is |ν(0) = |νe = |ν1 cos θ + |ν2 sin θ. Therefore, we have at time t |ν(t) = |ν1 cos θ e−iE1 t/¯h + |ν2 sin θ e−iE2 t/¯h . 1.1.2. The probability to ﬁnd this neutrino in the state |νe at time t is 2 Pe (t) = |νe |ν(t)|2 = cos2 θ e−iE1 t/¯h + sin2 θ e−iE2 t/¯h , which gives, after a simple calculation:

Pe (t) = 1 − sin2 (2θ) sin2

(E1 − E2 )t 2¯h

.

We have E1 − E2 = (m21 − m22 )c4 /(2pc). Deﬁning the oscillation length by L = 4π¯ hp/(|∆m2 | c2 ), we obtain

πct Pe (t) = 1 − sin2 (2θ) sin2 . L 1.1.3. For an energy E = pc = 4 MeV and a mass diﬀerence ∆m2 c4 = 10−4 eV2 , we obtain an oscillation length L = 100 km. 1.1.4. The time of ﬂight is t = /c. The probability Pe () is therefore

π 2 2 Pe () = 1 − sin (2θ) sin . (1.11) L

24

1 Neutrino Oscillations

1.1.5. A νµ energy of only 4 MeV is below the threshold of the reaction νµ + n → p + µ. Therefore this reaction does not occur with reactor neutrinos, and one cannot measure the νµ ﬂux. 1.1.6. In order to detect a signiﬁcant decrease in the neutrino ﬂux νe , we must have

π sin2 (2θ) sin2 > 0.1 . L (a) For the maximum mixing θ = π/4, i.e. sin2 (2θ) = 1, this implies π/L > 0.32 or > L/10. For E = 4 MeV and ∆m2 c4 = 10−4 eV2 , one ﬁnds > 10 km. The typical distances necessary to observe this phenomenon are of the order of a fraction of the oscillation length. (b) If the mixing is not maximum, one must operate at distances greater than L/10. Note that is the mixing angle is too small, (sin2 (2θ) < 0.1 i.e. θ < π/10), the oscillation amplitude is too weak to be detected, whatever the distance . In that case, one must improve the detection eﬃciency to obtain a positive conclusion. 1.1.7. (a) In all experiments except KamLAND, the distance is smaller than 1 km. Therefore, in all of these experiments |1 − Pe | ≤ 10−3 . The oscillation eﬀect is not detectable if the estimate |∆m2 | c4 ∼ 10−4 eV2 is correct. (b) For |∆m2 | c4 = 7.1 × 10−5 eV2 , tan2 θ = 0.45 and = 180 km, we obtain Pe = 0.50 which agrees with the measurement. The theoretical prediction taking into account the eﬀects due to the dispersion in energy is drawn on Fig. 1.3. We see incidentally how important it is to control error bars in such an experiment.

Ndetected N expected

1,2 1,0 0,8 0,6

ILL Bugey Rovno Goesgen Chooz KamLAND

0,4 0,2 0 10

102

distance (meters) 103

104

105

Fig. 1.3. Experimental points of Fig. 1.1 and the theoretical prediction of (1.11) (sinusoidal function damped by energy dispersion aﬀects). This curve is a best ﬁt of solar neutrino data. We notice that the KamLAND data point corresponds to the second oscillation of the curve

1.3 Solutions

25

Section 1.2: Oscillations of Three Species: Atmospheric Neutrinos 1.2.1. At time t = 0, we have: |ν(0) = |να =

Uαj |νj ,

j

and therefore at time t: |ν(t) = e−ipct/¯h

2 3

Uαj e−imj c

t/(2¯ hp)

|νj .

j

We conclude that the probability Pα→β to observe a neutrino of ﬂavor β at time t is 2 2 3 2 ∗ −imj c t/(2¯ hp) Uβj Uαj e Pα→β (t) = |νβ |ν(t)| = . j 1.2.2. We have Lij = 4π¯ hE/(|∆m2ij | c3 ). The oscillation lengths are proportional to the energy. We can use the result of question 1.3, with a conversion factor of 1000 to go from 4 MeV to 4 GeV. • For |∆m212 | c4 = 7.1 × 10−5 eV2 , we ﬁnd L12 = 140 000 km. • For |∆m223 | c4 = 2.5 × 10−3 eV2 , we ﬁnd L23 = 4 000 km. 1.2.3. We want to know the minimal distance necessary in order to observe oscillations. We assume that both mixing angles θ12 and θ23 are equal to π/4, which corresponds to maximum mixing. We saw in the ﬁrst part that if this mixing is not maximum, the visibility of the oscillations is reduced and that the distance which is necessary to observe the oscillation phenomenon is increased. By resuming the argument of the ﬁrst part, we ﬁnd that the modiﬁcation of the neutrino ﬂux of a given species is detectable beyond a distance ij such that sin2 (πij /Lij ) ≥ 0.1 i.e. ij ≥ Lij /10. This corresponds to 12 ≥ 14000 km for the oscillation resulting from the superposition 1 ↔ 2, and 23 ≥ 400 km for the oscillation resulting from the superposition 2 ↔ 3. 1.2.4. The factor of 2 between the expected muon and electron neutrino ﬂuxes comes from a simple counting. Each particle π − (resp. π + ) gives rise to a νµ , a ν¯µ and a ν¯e (resp. a νµ , a ν¯µ and a νe ). In practice, part of the muons reach the ground before decaying, which modiﬁes this ratio. Naturally, this eﬀect is taken into account in an accurate treatment of the data. 1.2.5. For an energy of 4 GeV, we have found that the minimum distance to observe the oscillation resulting from the 1 ↔ 2 superposition is 14000 km. We therefore remark that the oscillations νe νµ , corresponding to the mixing 1 ↔ 2 which we studied in the ﬁrst part cannot be observed at terrestrial distances. At such energies (4 GeV) and for evolution times corresponding at

26

1 Neutrino Oscillations

most to the diameter of the Earth (0.04 s), the energy diﬀerence E1 − E2 and the oscillations that it induces can be neglected. However, if the estimate |∆m223 | c4 > 10−3 eV2 is correct, the terrestrial distance scales allow in principle to observe oscillations resulting from 2 ↔ 3 and 1 ↔ 3 superpositions, which correspond to νµ ντ or νe ντ . 1.2.6. The angular distribution (therefore the distribution in ) observed for the νe ’s does not show any deviation from the prediction made without any oscillation. However, there is a clear indication for νµ oscillations: there is a deﬁcit of muon neutrinos coming from below, i.e. those which have had a long time to evolve. The deﬁcit in muon neutrinos is not due to the oscillation νe νµ of the ﬁrst part. Indeed, we have seen in the previous question that this oscillation is negligible at time scales of interest. The experimental data of Fig. 1.2 conﬁrm this observation. The deﬁcit in muon neutrinos coming from below is not accompanied with an increase of electron neutrinos. The eﬀect can only be due to a νµ ντ oscillation.1 No oscillation νe ντ appears in the data. In the framework of the present model, this is interpreted as the signature of a very small (if not zero) θ13 mixing angle. 1.2.7. Going back to the probability (1.11) written in question 1.4, the probability for an atmospheric muon neutrino νµ to be detected as a νµ is:

P () = 1 − sin (2θ23 ) sin 2

2

π L23

,

(1.12)

where the averaging is performed on the energy distribution of the neutrino. If we measure the neutrino ﬂux coming from the top, we have L23 , which gives Ptop = 1. If the neutrino comes from the bottom, the term sin2 (π/L23 ) averages to 1/2 and we ﬁnd: Pbottom = 1 −

1 sin2 (2θ23 ) . 2

The experimental data indicate that for −1 ≤ cos α ≤ −0.5, Pbottom = 1/2. The distribution is very ﬂat at a value of 100 events, i.e. half of the top value (200 events). We deduce that sin2 (2θ23 ) = 1, i.e. θ23 = π/4 and a maximum mixing angle for νµ ντ . The results published by Super-Kamiokande fully agree with this analysis.

1

For completeness, physicists have also examined the possibility of a “sterile” neutrino oscillation, i.e. an oscillation with a neutrino which would have no detectable interaction with matter.

1.4 Comments

27

1.4 Comments The diﬃculty of such experiments comes from the smallness of the neutrino interaction cross sections with matter. The detectors are enormous water tanks, where about ten events per day are observed (for instance ν¯e + p → e+ + n). The “accuracy” of a detector comes mainly from the statistics, i.e. the total number of events observed. In 1998, the ﬁrst undoubted observation of the oscillation ντ νµ was announced in Japan by the Super-Kamiokande experiment Fukuda Y. et al., Phys. Rev. Lett. 81, 1562 (1998)). This experiment uses a detector containing 50 000 tons of water, inside which 11 500 photomutipliers detect the Cherenkov light of the electrons or muons produced. About 60 ντ ’s were also detected, but this ﬁgure is too small to give further information. An accelerator experiment conﬁrmed the results afterwards (K2K collaboration, Phys. Rev. Lett. 90, 041801 (2003)). The KamLAND experiment is a collaboration between Japanese, American and Chinese physicists. The detector is a 1000 m3 volume ﬁlled with liquid scintillator (an organic liquid with global formula C-H). The name means KAMioka Liquid scintillator Anti-Neutrino Detector. Reference: KamLAND Collaboration, Phys. Rev. Lett. 90, 021802 (2003); see also http:/kamland.lbl.gov/. Very many experimental results come from solar neutrinos, which we have not dealt with here. This problem is extremely important, but somewhat too complex for our purpose. The pioneering work is due to Davis in his celebrated paper of 1964 (R. Davis Jr., Phys. Rev Lett. 13, 303 (1964)). Davis operated on a 37 Cl perchlorethylene detector and counted the number of 37 Ar atoms produced. In 25 years, his overall statistics has been 2200 events, i.e. one atom every 3 days! In 1991, the SAGE experiment done with Gallium conﬁrmed the deﬁcit (A. I. Abasov et al., Phys. Rev Lett. 67, 3332 (1991) and J. N. Abdurashitov et al., Phys. Rev Lett. 83, 4686 (1999)). In 1992, the GALLEX experiment, using a Gallium target in the Gran Sasso, also conﬁrmed the solar neutrino deﬁcit (P. Anselmann et al., Phys. Lett. B285, 376 (1992)). In 2001 the Sudbury Neutrino Observatory (SNO) gave decisive experimental results on solar neutrinos (Q.R. Ahmad et al., Phys. Rev. Lett. 87, 071307 (2001) and 89, 011301 (2002); see also M.B. Smy, Mod. Phys. Lett. A 17, 2163 (2002)). The 2002 Nobel prize for physics was awarded to Raymond Davis Jr. and Masatoshi Koshiba, who are the pioneers of this chapter of neutrino physics.

11 The EPR Problem and Bell’s Inequality

When a quantum system possesses more than one degree of freedom, the associated Hilbert space is a tensor product of the spaces associated to each degree of freedom. This structure leads to speciﬁc properties of quantum mechanics, whose paradoxical character has been pointed out by Einstein, Podolsky and Rosen. Here we study an example of such a situation, by considering entangled states for the spins of two particles. The system under consideration is a hydrogen atom which is dissociated into an electron and a proton. We consider the spin states of these two particles when they have left the dissociation region and are located in geometrically distinct regions, e.g. a few meters from one another. They are then free particles whose spin states do not evolve.

11.1 The Electron Spin Consider a unit vector uϕ in the (z, x) plane: uϕ = cos ϕ uz + sin ϕ ux , where ˆ e .uϕ the uz and ux are unit vectors along the z and x axes. We note Sˆeϕ = S component of the electron spin along the uϕ axis. 11.1.1. What are the eigenvalues of Sˆeϕ ? 11.1.2. We denote the eigenvectors of Sˆeϕ by |e : + ϕ and |e : − ϕ which, in the limit ϕ = 0, reduce respectively to the eigenvectors |e : + and |e : − of Sˆez . Express |e : + ϕ and |e : − ϕ in terms of |e : + and |e : −. 11.1.3. Assume the electron is emitted in the state |e : + ϕ. One measures the component Sˆeα of the spin along the direction uα = cos α uz + sin α ux . What is the probability P+ (α) of ﬁnding the electron in the state |e : + α? What is the expectation value Sˆeα in the spin state |e : + ϕ?

100

11 The EPR Problem and Bell’s Inequality

11.2 Correlations Between the Two Spins We ﬁrst assume that, after the dissociation, the electron–proton system is in the factorized spin state |e : + ϕ ⊗ |p : − ϕ. We recall that if |u1 and |u2 are vectors of E, and |v1 and |v2 of F , if ˆ act |u ⊗ |v belongs to the tensor product G = E ⊗ F , and if Aˆ and B ˆ acting in G, one has: respectively in E and F , Cˆ = Aˆ ⊗ B ˆ 1 v2 |B|v ˆ 1 . u2 | ⊗ v2 | Cˆ |u1 ⊗ |v1 = u2 |A|u

11.2.1. What is the probability P+ (α) of ﬁnding +¯ h/2 when measuring the component Sˆeα of the electron spin in this state? Having found this value, what is the state of the system after the measurement? Is the proton spin state aﬀected by the measurement of the electron spin? 11.2.2. Calculate the expectation values Sˆeα and Sˆpβ of the components of the electron and the proton spins along axes deﬁned respectively by uα and uβ = cos β uz + sin β ux . 11.2.3. The correlation coeﬃcient between the two spins E(α, β) is deﬁned as Sˆeα ⊗ Sˆpβ − Sˆeα Sˆpβ E(α, β) = . (11.1) 1/2 2 2 ˆ ˆ Seα Spβ Calculate E(α, β) in the state under consideration.

11.3 Correlations in the Singlet State We now assume that, after the dissociation, the two particles are in the singlet spin state: 1 |Ψs = √ |e : + ⊗ |p : − − |e : − ⊗ |p : + . (11.2) 2 11.3.1. One measures the component Sˆeα of the electron spin along the direction uα . Give the possible results and their probabilities. 11.3.2. Suppose the result of this measurement is +¯h/2. Later on, one measures the component Sˆpβ of the proton spin along the direction uβ . Here again give the possible results and their probabilities. 11.3.3. Would one have the same probabilities if the proton spin had been measured before the electron spin? Why was this result shocking for Einstein who claimed that “the real states of two spatially separated objects must be independent of one another”?

11.4 A Simple Hidden Variable Model

101

11.3.4. Calculate the expectation values Sˆeα and Sˆpβ of the electron and the proton spin components if the system is in the singlet state (11.2). 11.3.5. Calculate E(α, β) in the singlet state.

11.4 A Simple Hidden Variable Model For Einstein and several other physicists, the solution to the “paradox” uncovered in the previous section could come from the fact that the states of quantum mechanics, in particular the singlet state (11.2), provide an incomplete description of reality. A “complete” theory (for predicting spin measurements, in the present case) should incorporate additional variables or parameters, whose knowledge would render measurements independent for two spatially separated objects. However, present experiments cannot determine the values of these parameters, which are therefore called “hidden variables”. The experimental result should then consist in some averaging over these unknown parameters. In the case of interest, a very simpliﬁed example of such a theory is the following. We assume that, after each dissociation, the system is in a factorized state |e : +ϕ ⊗ |p : −ϕ, but that the direction ϕ varies from one event to the other. In this case, ϕ is the hidden variable. We assume that all directions ϕ are equally probable, i.e. the probability density that the decay occurs with direction ϕ is uniform and equal to 1/2π. Owing to this ignorance of the value of ϕ, the expectation value of an observable Aˆ is now deﬁned to be:

ˆ = A

1 2π

2π

e : + ϕ| ⊗ p : − ϕ| Aˆ |e : + ϕ ⊗ |p : − ϕ dϕ .

(11.3)

0

11.4.1. Using the deﬁnition (11.1) for E(α, β) and the new deﬁnition (11.3) for expectation values, calculate E(α, β) in this new theory. Compare the result with the one found using “orthodox” quantum mechanics in Sect. 3.5. 11.4.2. The ﬁrst precise experimental tests of hidden variable descriptions vs. quantum mechanics have been performed on correlated pairs of photons emitted in an atomic cascade.1 Although one is not dealing with spin-1/2 particles in this case, the physical content is basically the same as here. As an example, Fig. 11.1 presents experimental results obtained by A. Aspect and his collaborators in 1982. It gives the variation of E(α, β) as a function of the diﬀerence α − β, which is found to be the only experimentally relevant 1

The precision has now been greatly improved with the use of photon pairs produced by nonlinear splitting of ultraviolet photons (for a review, see e.g. A. Aspect, Nature, vol. 398, p. 189 (18 March 1999)).

102

11 The EPR Problem and Bell’s Inequality

quantity. Which theory, quantum mechanics or the simple hidden variable model developed above, gives a good account of the experimental data?

11.5 Bell’s Theorem and Experimental Results As proved by Bell in 1965, the disagreement between the predictions of quantum mechanics and those of hidden variable theories is actually very general when one considers correlation measurements on entangled states. We now show that the correlation results for hidden variable theories are constrained by what is known as Bell’s inequality, which, however, can be violated by quantum mechanics in speciﬁc experimental conﬁgurations. Consider a hidden variable theory, whose result consists in two functions A(λ, uα ) and B(λ, uβ ) giving respectively the results of the electron and proton spin measurements. Each of these two functions takes only the two values h/2 and −¯h/2. It depends on the value of the hidden variable λ for the con¯ sidered electron–proton pair. The nature of this hidden variable need not be further speciﬁed for the proof of Bell’s theorem. The result A of course depends on the axis uα chosen for the measurement of the electron spin, but it does not depend on the axis uβ . Similarly B does not depend on uα . This locality hypothesis is essential for the following discussion.

Fig. 11.1. Measured variation of E(α, β) as a function of α − β. The vertical bars represent the experimental error bars

11.6 Solutions

103

11.5.1. Give the correlation coeﬃcient E(α, β) for a hidden variable theory in terms of the functions A and B and the (unknown) distribution law P (λ) for the hidden variable λ. 11.5.2. Show that for any set uα , uα , uβ , uβ , one has A(λ, uα ) B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) − A(λ, uα ) B(λ, uβ ) = ±

¯2 h . 2

(11.4)

11.5.3. We deﬁne the quantity S as S = E(α, β) + E(α, β ) + E(α , β ) − E(α , β) . Derive Bell’s inequality |S| ≤ 2 . 11.5.4. Consider the particular case α − β = β − α = α − β = π/4, and compare the predictions of quantum mechanics with the constraint imposed by Bell’s inequality. 11.5.5. The experimental results obtained by A. Aspect et al. are E(α, β) = −0.66 (±0.04) for α−β = π/4 and E(α, β) = +0.68 (±0.03) for α−β = 3π/4. Is a description of these experimental results by a local hidden variable theory possible? Are these results compatible with quantum mechanics?

11.6 Solutions Section 11.1: The Electron Spin 11.1.1. In the eigenbasis |e : ± of Sˆez , the matrix of Sˆeϕ is

h cos ϕ ¯ sin ϕ . 2 sin ϕ − cos ϕ The eigenvalues of this operator are +¯h/2 and −¯ h/2. 11.1.2. The corresponding eigenvectors are ϕ ϕ |e : + ϕ = cos |e : + + sin |e : − 2 2 ϕ ϕ |e : − ϕ = − sin |e : + + cos |e : −. 2 2 11.1.3. The probability amplitude is e : + α|e : + ϕ = cos((ϕ − α)/2) and the probability P+ (α) = cos2 ((ϕ − α)/2). Similarly P− (α) = sin2 ((ϕ − α)/2), and the expectation value is, ﬁnally, Sˆeα =

¯ h cos (ϕ − α) . 2

104

11 The EPR Problem and Bell’s Inequality

Section 11.2: Correlations Between the Two Spins 11.2.1. The projector on the eigenstate |e : + α, corresponding to the measured value, is |e : + α e : + α| ⊗ Iˆp , where Iˆp is the identity operator on the proton states. Therefore P+ (α) = |e : + α|e : + ϕ|2 = cos2

ϕ−α , 2

and the state after measurement is |e : + α ⊗ |p : − ϕ. The proton spin is not aﬀected, because the initial state is factorized (all probability laws are factorized). 11.2.2. One has Sˆeα =

h ¯ 2

cos (ϕ − α) and Sˆpβ = − h¯2 cos (ϕ − β).

11.2.3. By deﬁnition, one has: ¯2 ˆ h 2 Sˆeα Ie = 4

and

¯2 ˆ h 2 Ip Sˆpβ = 4

and Sˆeα ⊗ Sˆpβ = e : + ϕ|Sˆeα |e : + ϕ p : − ϕ|Sˆpβ |p : − ϕ =−

¯2 h cos(ϕ − α) cos(ϕ − β) . 4

Therefore E(α, β) = 0. This just reﬂects the fact that in a factorized state, the two spin variables are independent. Section 11.3: Correlations in the Singlet State 11.3.1. There are two possible values: h/2, corresponding to the projector |e : + α e : + α| ⊗ Iˆp , and ¯ −¯ h/2, corresponding to the projector |e : − α e : − α| ⊗ Iˆp . Therefore, the probabilities are 1 2 2 P+ (α) = |e : + α|e : +| + |e : + α|e : −| = 1/2 2 and similarly P− (α) = 1/2. This result is a consequence of the rotational invariance of the singlet state. 11.3.2. The state after the measurement of the electron spin, yielding the result +¯ h/2, is cos

α α |e : + α ⊗ |p : − − sin |e : + α ⊗ |p : + = |e : + α ⊗ |p : − α . 2 2

This simple result is also a consequence of the rotational invariance of the singlet state, which can be written as

11.6 Solutions

105

1 |Ψs = √ (|e : + α ⊗ |p : − α − |e : − α ⊗ |p : + α) . 2 Now the two possible results for the measurement of the proton spin ±¯ h/2 have probabilities P+ (β) = sin2

α−β 2

P− (β) = cos2

α−β . 2

11.3.3. If one had measured Sˆpβ ﬁrst, one would have found P+ (β) = P− (β) = 1/2. The fact that a measurement on the electron aﬀects the probabilities for the results of a measurement on the proton, although the two particles are spatially separated, is in contradiction with Einstein’s assertion, or belief. This is the starting point of the Einstein–Podolsky–Rosen paradox. Quantum mechanics is not a local theory as far as measurement is concerned. Note, however, that this non-locality does not allow the instantaneous transmission of information. From a single measurement of the proton spin, one cannot determine whether the electron spin has been previously measured. It is only when, for a series of experiments, the results of the measurements on the electron and the proton are later compared, that one can ﬁnd this non-local character of quantum mechanics. 11.3.4. Individually, the expectation values vanish, since one does not worry about the other variable: Sˆeα = Sˆpβ = 0 . 11.3.5. However, the spins are correlated and we have

α−β α−β h2 ¯ − cos2 Sˆeα ⊗ Sˆpβ = sin2 4 2 2 and therefore

E(α, β) = − cos(α − β).

Section 11.4: A Simple Hidden Variable Model 11.4.1. Using the results of Sect. 2, we have: dϕ h ¯ cos(ϕ − α) =0 Sˆeα = 2 2π and similarly Sˆpβ = 0. We also obtain dϕ ¯2 h cos(ϕ − α) cos(ϕ − β) 4 2π 2 h ¯ = − cos(α − β) . 8

Sˆeα ⊗ Sˆpβ = −

106

11 The EPR Problem and Bell’s Inequality

Therefore, in this simple hidden variable model, 1 E(α, β) = − cos(α − β). 2 In such a model, one ﬁnds a non-vanishing correlation coeﬃcient, which is an interesting observation. Even more interesting is that this correlation is smaller than the prediction of quantum mechanics by a factor 2. 11.4.2. The experimental points agree with the predictions of quantum mechanics, and undoubtedly disagree with the results of the particular hidden variable model we have considered. We must however point out that the data given in the text is not the actual measured data. The “true” results are shown in Fig. 11.2, where the error bars correspond only to statistical errors. The diﬀerence from theory (i.e. quantum mechanics) is due to systematic errors, mainly the acceptance of the detectors.

Fig. 11.2. Actual experimental variation of E(α, β) as a function of α − β

Section 11.5: Bell’s Theorem and Experimental Results 11.5.1. In the framework of a hidden variable theory, the correlation coeﬃcient is 4 P (λ) A(λ, uα ) B(λ, uβ ) dλ , E(α, β) = 2 h ¯ where P (λ) is the (unknown) distribution law for the variable λ, with P (λ) > 0 ∀λ and P (λ) dλ = 1 .

11.6 Solutions

107

Note that we assume here that the hidden variable theory reproduces the one-operator averages found for the singlet state: Spβ = P (λ) B(λ, uβ ) dλ = 0. Seα = P (λ) A(λ, uα ) dλ = 0 If this was not the case, such a hidden variable theory should clearly be rejected since it would not reproduce a well established experimental result. 11.5.2. The quantity of interest can be written:

A(λ, uα ) B(λ, uβ ) + B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) − B(λ, uβ ) . h/2. The two quantities B(λ, uβ ) and B(λ, uβ ) can take only the two values ±¯ Therefore one has either h B(λ, uβ ) + B(λ, uβ ) = ±¯ or

B(λ, uβ ) + B(λ, uβ ) = 0

B(λ, uβ ) − B(λ, uβ ) = 0 B(λ, uβ ) − B(λ, uβ ) = ±¯ h,

¯ /2. hence the result, since |A(λ, uα )| = |A(λ, uβ )| = h 11.5.3. We multiply the result (11.4) by P (λ) and integrate over λ. Bell’s inequality follows immediately. 11.5.4. The quantum mechanical result for S is SQ = − cos(α − β) − cos(α − β ) − cos(α − β ) + cos(α − β) . In general, if we set θ1 = α − β, θ2 = β − α, θ3 = α − β , we can look for the extrema of f (θ1 , θ2 , θ3 ) = cos(θ1 + θ2 + θ3 ) − (cos θ1 + cos θ2 + cos θ3 ) . The extrema correspond to θ1 = θ2 = θ3 and sin θ1 = sin 3θ1 , whose solutions between 0 and π are θ1 = 0, π/4, 3π/4, π . Deﬁning the √ function g(θ1 )√= −3 cos θ1 + cos 3θ1 we have: g(0) = −2, g(π/4) = −2 2, g(3π/4) = 2 2, g(π) = 2. We have represented the variation of g(θ) in Fig. 11.3. The shaded areas correspond to results which cannot be explained by hidden variable√theories. In particular, for α − β = β − α = α − β = π/4, we get SQ = −2 2, which clearly violates Bell’s inequality. This system constitutes therefore a test of the predictions of quantum mechanics vs. any local hidden variable theory. 11.5.5. The numbers given in the text lead to |3E(π/4) − E(3π/4)| = √ 2.66 (±0.15) in excellent agreement with quantum mechanics (2 2) but incompatible with hidden variable theories. As in the previous question, the actual measurements were in fact E(π/4) = −0.62 (±0.04), E(3π/4) = 0.60 (±0.03), therefore |3E(π/4) − E(3π/4)| =

108

11 The EPR Problem and Bell’s Inequality

2.46 (±0.15) which violates unquestionably Bell’s inequality, and is consistent with the quantum mechanical prediction. It is therefore not possible to ﬁnd a local hidden variable theory which gives a good account of experiment.

Fig. 11.3. Variation of g(θ), as deﬁned in the text

Section 11.6: References A. Einstein, B. Podolsky, and N. Rosen, Phys. Rev. 47, 777 (1935). J.S. Bell, Physics 1, 195 (1964); see also J. Bell, Speakable and unspeakable in quantum mechanics, Cambridge University Press, Cambridge (1993). The experimental data shown here are taken from: A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1982); A. Aspect, J. Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804 (1982).

18 Exact Results for the Three-Body Problem

The three-body problem is a famous question of mechanics. Henri Poincar´e was the ﬁrst to prove exact properties, and this contributed to his celebrity. The purpose of this chapter is to derive some rigorous results for the threebody problem in quantum mechanics. Here we are interested in obtaining rigorous lower bounds on three-body ground state energies. Upper bounds are easier to obtain by variational calculations. We will see that our lower bounds are actually quite close to the exact answers, to which they provide useful approximations.

18.1 The Two-Body Problem Consider a system of two particles with equal masses m and momenta p1 and p2 , interacting via a potential V (r12 ) where r12 = |r 1 − r 2 |. ˆ of the system. Let P = p1 + p2 and p = 18.1.1. Write the Hamiltonian H (p1 − p2 )/2 be the total and relative momentum. ˆ 12 Hamiltonians by writing ˆ cm and the relative H Separate the center of mass H ˆ as: H ˆ =H ˆ cm + H ˆ 12 , H

Hcm

ˆ2 P , = 2M

ˆ2 ˆ 12 = p H + V (ˆ r12 ) , 2µ

(18.1)

where M = 2m is the total mass of the system. Give the value of the reduced mass µ in terms of m. ˆ 12 . Give the ex18.1.2. We denote by E (2) (µ) the ground state energy of H (2) 2 pression for E (µ) in the two cases V (r) = −b /r and V (r) = κr2 /2.

186

18 Exact Results for the Three-Body Problem

18.2 The Variational Method ˆ and {En } the Let {|n} be the orthonormal eigenstates of a Hamiltonian H ordered sequence of its corresponding eigenvalues: E0 < E1 < E2 < · · · . ˆ 18.2.1. Show that n|H|n = En . 18.2.2. Consider an arbitrary vector |ψ of the Hilbert space of the system. By expanding |ψ on the basis {|n}, prove the inequality ∀ψ ,

ˆ ψ|H|ψ ≥ E0 ψ|ψ.

(18.2)

ˆ is the Hamiltonian 18.2.3. Show that the previous result remains valid if H of a two-body subsystem and |ψ a three-body state. In order to do so, one ˆ 12 the Hamiltonian of the (1, 2) subsystem in the three-body can denote by H system of wave function ψ(r 1 , r 2 , r 3 ). One can ﬁrst consider a given value of r 3 , and then integrate the result over this variable.

18.3 Relating the Three-Body and Two-Body Sectors Consider a system of three-particles of equal masses m with pairwise interactions: V = V (r12 ) + V (r13 ) + V (r23 ) . 18.3.1. Check the identity 2

2

2

3(p21 + p22 + p23 ) = (p1 + p2 + p3 ) + (p1 − p2 ) + (p2 − p3 ) + (p3 − p1 )

2

ˆ (3) can be written as and show that the three-body Hamiltonian H ˆ (3) = H ˆ cm + H ˆ (3) , H rel

ˆ2 ˆ cm = P , H 6m

ˆ =p ˆ1 + p ˆ2 + p ˆ 3 is the total three-body momentum, and where the where P ˆ (3) is a sum of two-particle Hamiltonians of the type relative Hamiltonian H rel deﬁned in (18.1), ˆ 12 + H ˆ 23 + H ˆ 31 ˆ (3) = H H rel with a new value µ of the reduced mass. Express µ in terms of m. ˆ ij commute in general? What would 18.3.2. Do the two-body Hamiltonians H be the result if they did? ˆ (3) , and E (3) the cor18.3.3. We call |Ω the normalized ground state of H rel responding energy. Show that the three-body ground state energy is related to the ground state energy of each two-body subsystem by the inequality: E (3) ≥ 3E (2) (µ ).

(18.3)

18.5 From Mesons to Baryons in the Quark Model

187

18.3.4. Which lower bounds on the three-body ground-state energy E (3) does one obtain in the two cases V (r) = −b2 /r and V (r) = κr2 /2? In the ﬁrst case, the exact result, which can be obtained numerically, is E (3) h2 . How does this compare with the bound (18.3)? −1.067 mb4 /¯

18.4 The Three-Body Harmonic Oscillator The three-body problem can be solved exactly in the case of harmonic interactions V (r) = κr2 /2. In order to do this, we introduce the Jacobi variables: √ √ √ ˆ 2 = (2ˆ ˆ 3 = (ˆ ˆ 1 = (ˆ r 1 − rˆ 2 )/ 2, R r 3 − rˆ 1 − rˆ 2 )/ 6, R r 1 + rˆ 2 + rˆ 3 )/ 3 R √ √ √ ˆ = (2ˆ ˆ = (ˆ ˆ = (ˆ ˆ 2 )/ 2, Q ˆ1 − p ˆ 2 )/ 6, Q ˆ2 + p ˆ 3 )/ 3 . p1 − p p3 − p p1 + p Q 1 2 3 ˆα 18.4.1. What are the commutation relations between the components R j ˆ j and Q ˆ k , (α = 1, 2, 3, and β = 1, 2, 3)? ˆ β of R and Q k 18.4.2. Check that one has Q21 + Q22 + Q23 = p21 + p22 + p23 , and: 2

2

2

3(R12 + R22 ) = (r 1 − r 2 ) + (r 2 − r 3 ) + (r 3 − r 1 ) . 18.4.3. Rewrite the three-body Hamiltonian in terms of these variables for a harmonic two-body interaction V (r) = κr2 /2. Derive the three-body ground state energy from the result. Show that the inequality (18.3) is saturated, i.e. the bound (18.3) coincides with the exact result in that case. Do you think that the bound (18.3), which is valid for any potential, can be improved without further specifying the potential?

18.5 From Mesons to Baryons in the Quark Model In elementary particle physics, the previous results are of particular interest since mesons are bound states of two quarks, whereas baryons, such as the proton, are bound states of three quarks. Furthermore, it is an empirical observation that the spectroscopy of mesons and baryons is very well accounted for by non-relativistic potential models for systems of quarks. The φ meson, for instance, is a bound state of a strange quark s and its antiquark s¯, both of same mass ms . The mass mφ is given by mφ = 2ms + E (2) (µ)/c2 where µ = ms /2, c is the velocity of light, and E (2) is the ground state energy of the s¯ s system which is bound by a potential Vqq¯(r). The Ω − baryon is made of three strange quarks. Its mass is given by MΩ = 3ms + E (3) /c2 , where E (3) is the ground state energy of the three s quarks, which interact pairwise through a two-body potential Vqq (r). These potentials are related very simply to each other by

188

18 Exact Results for the Three-Body Problem

Vqq (r) =

1 Vqq¯(r) . 2

It is a remarkable property, called ﬂavor independence, that these potentials are the same for all types of quarks. 18.5.1. Following a procedure similar to that of Sect. 3, show that E (3) ≥ (3/2)E (2) (µ ); express µ in terms of µ = ms /2. 18.5.2. Consider the potential Vqq¯(r) = g ln(r/r0 ), and the two-body Hamilˆ (2) (˜ ˆ (2) (µ) and H µ) corresponding to the same potential but diﬀerent tonians H ˆ (2) (µ) + C, ˆ (2) (˜ µ) into H reduced masses µ and µ ˜. By rescaling r, transform H where C is a constant. (2) ˆ (2) (µ) Calculate the value of C and show that the eigenvalues En (µ) of H (2) ˆ (2) (˜ and En (˜ µ) of H µ) are related by the simple formula µ) = En(2) (µ) + En(2) (˜

g µ ln . 2 µ ˜

18.5.3. A striking characteristic of the level spacings in quark–antiquark systems is that these spacings are approximately independent of the nature of the quarks under consideration, therefore independent of the quark masses. Why does this justify the form of the above potential Vqq¯(r) = g ln(r/r0 )? 18.5.4. Show that the following relation holds between the Ω − and φ masses MΩ and mφ : 3 M Ω ≥ mφ + a 2 and express the constant a in terms of the coupling constant g. 18.5.5. The observed masses are mφ = 1019 MeV/c2 and MΩ = 1672 MeV/c2 . The coupling constant is g = 650 MeV. Test the inequality with these data.

18.6 Solutions Section 18.1: The Two-Body Problem 18.1.1. The two-body Hamiltonian is ˆ 21 ˆ2 p ˆ = p H + 2 + Vˆ (r12 ) . 2m 2m The center of mass motion can be separated as usual: ˆ2 ˆ2 ˆ = P +p H + Vˆ (r12 ) , 2M 2µ where M = 2m and µ = m/2 are respectively the total mass and the reduced mass of the system.

18.6 Solutions

189

18.1.2. For a Coulomb-type interaction V (r) = −b2 /r, we get E (2) (µ) = −

µb4 . 2¯ h2

For a harmonic interaction V (r) = κr2 /2, we get κ 3 (2) E (µ) = ¯ h . 2 µ Section 18.2: The Variational Method ˆ 18.2.1. By deﬁnition, n|H|n = En n|n = En . 18.2.2.Since {|n} is a basis of the Hilbert space, |ψcan be expanded as |ψ = cn |n, and of its norm is ψ|ψ = |cn |2 . We therefore the square 2 ˆ have ψ|H|ψ = En |cn | . If we simply write ˆ (En − E0 )|cn |2 , ψ|H|ψ − E0 ψ|ψ = we obtain, since En ≥ E0 and |cn |2 ≥ 0: ˆ ψ|H|ψ ≥ E0 ψ|ψ . ˆ =H ˆ 12 , for ﬁxed r 3 , ψ(r 1 , r 2 , r 3 ) can be considered as a non18.2.3. If H normalized two-body wave function. Therefore ∗ ˆ 12 ψ(r 1 , r 2 , r 3 ) d3 r1 d3 r2 ψ (r 1 , r 2 , r 3 ) H ≥ E0 |ψ(r 1 , r 2 , r 3 )|2 d3 r1 d3 r2 . By integrating this inequality over r 3 , one obtains the desired result. Section 18.3: Relating the Three-Body and Two-Body Sectors 18.3.1. The identity is obvious, since the crossed terms vanish on the rightˆ 2 /(6m) + H ˆ =P ˆ 12 + H ˆ 23 + H ˆ 31 , with hand side. Therefore H

2

#2 " ˆj ˆ j /2 ˆi − p ˆi − p p p ˆ ˆ Hij = + V (rij ) = + Vˆ (rij ) (18.4) 6m 2µ with a reduced mass µ = 3m/4. ˆ 12 and H ˆ 23 do not commute; for instance p ˆ1 − p ˆ 2 does 18.3.2. Obviously, H ˆ not commute with V (r23 ). If they did, the three-body energies would just be the sum of two-body energies as calculated with a reduced mass µ = 3m/4, and the solution of the three-body problem would be simple.

190

18 Exact Results for the Three-Body Problem

ˆ (3) |Ω = Ω|H ˆ ij |Ω. However, owing 18.3.3. By deﬁnition, E (3) = Ω|H rel ˆ ij |Ω ≥ E (2) (µ ), so that to the results of questions 2.2 and 2.3, we have Ω|H E (3) ≥ 3E (2) (µ )

with

µ = 3m/4 .

18.3.4. For a Coulomb-type potential, we obtain E (3) ≥ −

3 µ b4 9 mb4 =− , 2 2 ¯ 8 ¯h2 h

which deviates by only 6% from the exact answer −1.067 mb4 /¯ h2 . In the harmonic case, we obtain: √ κ κ 3 (3) h . E ≥3 ¯ = 3 3 ¯h 2 µ m Section 18.4: The Three-Body Harmonic Oscillator 18.4.1. One easily veriﬁes that Jacobi variables satisfy canonical commutation relations: ˆα, Q ˆ β ] = i¯ [R h δjk δαβ . j k 18.4.2. These relations are a simple algebraic exercise. 18.4.3. We ﬁnd ˆ2 ˆ2 ˆ2 ˆ = Q1 + 3 κR ˆ 2 + Q2 + 3 κR ˆ 2 + Q3 = H ˆ1 + H ˆ2 + H ˆ cm , H 1 2 2m 2 2m 2 2m ˆ cm = Q ˆ 2 /(2m) = Pˆ 2 /(6m) is the center of mass Hamiltonian. The where H 3 ˆ 2 , and H ˆ cm commute. The ground state energy (with ˆ 1, H three Hamiltonians H the center of mass at rest) is therefore √ 3κ κ 3 (3) h = 3 3 ¯h , E =2 ¯ 2 m m which coincides with the lower bound obtained in question 3.4. The bound is therefore saturated if the interaction is harmonic. In order to improve the bound, one must further specify the interaction. Actually, the bound is saturated if and only if the interaction potential is harmonic. Indeed the variational inequality we use becomes an equality if and only if the wave function coincides with the exact ground state wave function. Owing to the particular symmetry of quadratic forms, the Jacobi variables guarantee that this happens in the harmonic case. The property ceases to be true for any other potential.

18.6 Solutions

191

Section 18.5: From Mesons to Baryons in the Quark Model 18.5.1. The s¯ s relative Hamiltonian is 2 ˆ (2) = pˆ + Vqq¯(ˆ H r) . ms

The sss relative Hamiltonian is (cf. Sect. 3): (ˆ pi − pˆj )2 1 (3) ˆ + Vqq¯(ˆ rij ) H = 6ms 2 i<j

pi − pˆj )2 1 (ˆ = + Vqq¯(ˆ rij ) . 2 i<j 3ms

(18.5) (18.6)

Therefore, ˆ (3) = 2H

2

ˆ ij H

pi − pˆj ) /2) ˆ ij = ((ˆ H + Vqq¯(ˆ rij ) 2µ

with

i<j

with µ = 3ms /8 = 3µ/4. From this relation we deduce the inequality: 2 E (3) ≥ 3 E (2) (µ )

µ = 3µ/4 .

with

18.5.2. With the rescaling r → αr, one obtains: ˆ (2) (˜ H µ) = The choice α =

r pˆ2 + g ln + g ln α . 2 2α µ ˜ r0

ˆ (2) (˜ ˆ (2) (µ) + g ln α so that µ/˜ µ leads to H µ) = H En(2) (˜ µ) = En(2) (µ) +

g µ ln . 2 µ ˜

18.5.3. In a logarithmic potential, the level spacing is independent of the mass. This is a remarkable feature of the observed spectra, at least for heavy quarks, and justiﬁes the investigation of the logarithmic potential. Amazingly enough, this empirical prescription works quite well for light quarks, although one might expect that a relativistic treatment is necessary. 18.5.4. The binding energies satisfy

3 g 4 E (3) ≥ E (2) + ln 2 2 3 with MΩ = 3ms +

E (3) c2

We therefore obtain MΩ ≥

mφ = 2ms +

3 3g 4 mφ + 2 ln . 2 4c 3

E (2) . c2

192

18 Exact Results for the Three-Body Problem

18.5.5. For g = 650 MeV and a = 140 MeV/c2 , we obtain MΩ c2 = 1672 MeV ≥ 1669 MeV , which is remarkably accurate. Actually, the quark–quark potential is only logarithmic at distances smaller than 1 fm, which corresponds to the φ mean square radius. At larger distances, it grows more rapidly (linearly). Such inequalities are quite useful in practice for deciding what choice to make for the potential and for its domain of validity. The generalization of such inequalities can be found in the literature quoted below. They are useful in a variety of physical problems. References J.-L. Basdevant, J.-M. Richard, and A. Martin, Nuclear Physics B343, 60, 69 (1990). J-L. Basdevant, J.-M. Richard, A. Martin, and Tai Tsun Wu, Nuclear Physics B393, 111 (1993).

Author Index

Abasov, A.I. et al., 27 Abdurashitov, J.N. et al., 27 Ahmad, Q. R. et al., 27 Anderson, M.H., 202 Andrews, M.R., 202 Anselmann, P. et al., 27 Ashkin, A., 276 Aspect, A., 101, 108 Barnett, S.M., 167 Basdevant, J.-L., 192 Bastard, G., VII Bell, J.S., 99, 108 Ben Dahan, M., 285 Berkhout, J., 265 Berko, S., 80 Berlinsky, A.J., 265 Biraben, F., 82 Bjorkholm, J.E., 276 Bouchoule, I., 285 Bradley, C., 202 Brune, M., 119, 145 Cable, A., 276 Castin, Y., VII, 285 Chazalviel, J.-N., VII Chazalviel, J.N., 239 Chu, S., 80, 276 Clairon, A., 36 Cohen-Tannoudji, C., 276 Collela, A.R., 45 Corben, H.C., 80 Cornell, E.A., 202 Courty, J.-M., VII

Crane, H.R., 66 Dalibard, J., 108 Davis R., 27 DeBenedetti, S., 80 Dehmelt, H., 276 Delande, D., VII, 82 Dorner, B., 213 Doyle, J.M., 265 Dress, W.D., 53 Dreyer, J., 119, 145 Durfee, D.S., 202 Einstein, A., 99, 100, 108 Ensher, J.R., 202 Equer, B., VII Fischer, M., 36 Fukuda Y., 27 Gabrielse, G., 182 GALLEX Collaboration, 27 Gay, J.-C., 82 Gillet, V., VII Grangier, P., VII, 108, 153, 167 Greenberger, D., 45 Greene, G.L., 53 Grynberg, G., VII H¨ ansch, T.W., 276 Hagley, E., 119, 145 Hameau, S., 229 Haroche, S., 119, 145 Hollberg, L., 276

288

Author Index

Hulet, R.G., 202

Prodan, J.V., 276

Jacquet, F., VII Jolicoeur, T., VII K2K collaboration, 27 KamLAND collaboration, 27 Ketterle, W., 202 Kinoshita, T., 80 Koshiba M., 27 Kurn, D.M., 202

Raimond, J.-M., 119, 145 Raizen, M., 285 Ramsey, N.F., 47, 53, 156 Reichel, J., 285 Rich, J., VII Richard, J.-M., 192 Roger, G., 108 Rosen, N., 99, 108 Roug´e, A., VII

Maali, A., 119, 145 Maitre, X., 119 Maleki, L., 36 Mampe, W., 53 Marion, H., 36 Martin, A., 192 Matthews, M.R., 202 Metcalf, H., 276 Mewes, M.-O., 202 Miller, P.D., 53 Mills, A.P., 80

Sackett, C.A., 202 SAGE Collaboration, 27 Salomon, C., 36, 285 Santarelli, G., 36 Schawlow. A., 276 Schmidt-Kaler, F., 145 Schr¨ odinger, 109 Smith, J.H., 166 Smith, K.F., 53, 167 Smy, M.B., 27 Stoler, D., 119

Neuman, J. von, 147 Niu, Q., 285

Tjoelker, R.L., 36

Overhauser, A.W., 37, 45

van Druten, N.J., 202 von Neuman, J., 147

Peik, E., 285 Peil, S., 182 Pendlebury, J.M., 53 Pendleton, H.N., 80 Penent, F., 82 Perrin, P., 53 Phillips, W.D., 276 Platt, J.R., 235 Podolsky, B., 99, 108 Poizat, J.P., 153, 167 Prestage, J.D., 36

Weinheimer et al., 72 Werner, S.A., 45 Wieman, C.E., 202 Wilkinson, D.T., 66 Wineland, D., 276 Wu, Tai Tsun, 192 Wunderlich, C., 119 Yurke, B., 119 Zimmerman, D.S., 265

Subject Index

absorption, 239, 272 addition of angular momenta, 7, 29 adiabatic theorem, 278 alakali atom, 29 angular momenta addition, 7, 29 angular momentum, 6 annihilation, 75, 132 annihilation and creation operators, 109 antineutrino, 69 antinode, 234 antiparticle, 17 antiquark, 187 atmospheric neutrinos, 20 atomic clocks, 29 atomic fountain, 31 β decay, 17, 69 Band structure, 277 baryon, 187 Bell’s inequality, 99, 102, 107 Bell’s theorem, 102 Bethe-Bloch formula, 94 Bloch index, 278 Bloch oscillations, 277, 278, 284 Bloch theorem, 278 blue shift, 233 Bohr radius, 69, 82, 240 Born approximation, 12 Bose–Einstein condensate, 193, 195 boson, 10 Bragg angle, 37 Bragg reﬂection, 285

Brillouin zone, 281, 284 broad line condition, 273 cat paradox, 109 cesium, 29, 32 cesium atom, 283 cesium atoms, 279 • CH3 (methyl), 239 CH3 −• COH−COO− , 240 chain of coupled spins, 203 chain of molecules, 204 coherence properties, 193 collision, 12 condensate, 193 conductor, 285 cooling time, 273, 274 correlated pairs of spins, 122 correlation coeﬃcient, 100 Coulomb correlations, 229 Coulomb potential, 8 creation, 132 cross section, 12 crossed Kerr eﬀect, 153 cryptography, 121 CSCO, 3 Cs Fe Br3 , 204 damped oscillation, 135 damping, 113 damping time, 137, 144 decay, 75 degeneracy, 206 density operator, 3, 247, 251, 267 diﬀraction peak, 60

290

Subject Index

Dirac distribution, 5 dispersion relation, 257 Doppler cooling, 267, 269 Doppler shift, 52 drift of fundamental constants, 32

hydrogen hyperﬁne hyperﬁne hyperﬁne hyperﬁne

Ehrenfest theorem, 4, 59, 61, 67, 139 Einstein-Podolsky-Rosen paradox, 99 electric dipole, 267 electric dipole coupling, 277 electric dipole moment, 166 electromagnetic wave, 31 electron magnetic moment anomaly, 65 electron spin, 100 electron spin resonance, 237 energy bands, 281 energy loss, 87 entangled states, 3, 99 entanglement, 3 eraser, 155 excitons, 206

identical particles, 10 interaction time, 88, 91 interference, 155, 159 ion beams, 94 ion therapy, 94 ionization energy, 69 ions in matter, 87 isotope, 69

atom, 73, 81 interaction, 29, 237, 238 splitting, 73 structure, 237 transition, 74

Jacobi variables, 187, 190 joint probability, 165 KamLAND, 19 ket, 1

G.P.S., 29, 32 gravitational ﬁeld, 37 gravitational potential, 43 gyromagnetic ratios, 237

Laboratoire Kastler Brossel, 145 Laguerre polynomial, 9 Larmor frequency, 67 Larmor precession, 47 laser cooling, 267 LEP, 18 lepton, 17 lifetime, 12 liquid helium, 255, 258 lithium atoms, 197 local hidden variable theory, 103 local probe, 245 local theory, 105 locality, 102 logarithmic potential, 191 lower bounds, 185

Hamiltonian, 2 harmonic oscillator, 7, 109, 197 harmonic potential, 193 3 He+ , 69 heavy ion therapy, 94 Heisenberg inequalities, 60 hidden variables, 99, 101 Hilbert space, 1 holes, 211 hydrocarbon ions, 231 hydrogen, 249, 255

magnetic excitations, 203 magnetic excitons, 203, 206 magnetic interaction, 29, 41 magnetic moment, 58 magnetic resonance, 155 measurement, 147, 158 measuring apparatus, 63, 159 meson, 17, 187 modes, 267 molecular ions, 231 muon, 17

factorized spin, 100 Fermi golden rule, 11 fermion, 10 ﬁeld quantization, 131 ﬁne structure constant, 32 ﬂavor, 20 form factor, 13 Fourier transform, 4 free radicals, 237

Subject Index muon spin, 245 muonium, 245 muonium in silicon, 247

proton spin, 100, 105 proton therapy, 94 pure case, 1

n photon state, 132 neutrino, 17, 253 neutrino mass, 71 neutrino masses, 18 neutrino oscillations, 17 neutron, 55, 155, 206 neutron beam, 37, 47, 55 neutron gyromagnetic ratio, 49 neutron interference, 37 neutron magnetic moment, 59 nitrogenous ions, 232 node, 234 nuclear reaction, 87 nucleus, 29, 69

QND measurement, 153, 167 quantum boxes, 215 quantum cryptography, 121, 125 quantum dots, 215 quantum eraser, 155, 159 quantum measurement, 2 quantum paths, 155, 162 quantum reﬂection, 255 quantum revival, 137 quantum superposition, 112, 113 quark masses, 188 quark model, 187 quasi-classical states, 110, 133 quasi-particles, 203

observable, 2 Operator, 2 Operator energy, 5 momentum, 5 position, 5 optical Bloch equations, 267, 271 optronics, 229 orbital angular momentum, 6 oscillation Rabi, 11 oscillation lengths, 21

Rabi oscillation, 11, 131, 143 radiation pressure, 268 radioactive, 69 Ramsey fringes, 47, 156 red shift, 233 reﬂection, 255 relaxation, 267 resonance curve, 50 revival time, 144 ripplon, 255, 257, 263, 265 rubidium, 29, 32, 135

pairs of photons, 101 Pauli hydrogen atom, 81 Pauli matrices, 40, 121, 237 Pauli principle, 231 periodic potential, 277 perturbation theory, 9 phase states, 148 phonon, 229 photon pairs, 129 pigments, 231 polyethylene, 231 positive muon, 245 positron, 73, 253 positronium, 73 potential well, 231 precession, 65

satellite, 32 scalar product, 1 scattering, 12 scattering amplitude, 13 scattering length, 14, 194 scattering state, 13 Schr¨ odinger’s cat, 109 Schr¨ odinger equation, 2 semiconductor, 229 semiconductor detectors, 89 singlet state, 76, 127 sodium atom, 268, 269 sodium condensate, 196 spatial state, 57 spherical harmonics, 6 spin, 7 spin excitation wave, 207

291

292

Subject Index

spin state, 41, 47, 57, 122, 155, 158, 159, 246 spin transitions, 238 spin–spin interaction, 204, 246 spontaneous emission, 268, 272 standing laser wave, 277 Stark eﬀect, 81 state, 1 statistical mixture, 3, 112, 113 sterile neutrino, 26 Stern–Gerlach experiment, 55, 63 sticking probability, 255, 265 stimulated emission, 272 Super-Kamiokande, 21 superconducting niobium cavity, 145 supernovae, 18 superposition principle, 1 τ lepton, 17 Thomas–Reiche–Kuhn sum rule, 89 three-body harmonic oscillator, 187 three-body problem, 185 time–energy uncertainty relation, 63 time-dependent perturbations, 11 transition amplitude, 51

triplet states, 76, 77 tritium atom, 69 tritium decay, 69 uncertainty relation, 4, 5, 50 unitary operator, 277 unpaired electron, 237 Van der Waals potential, 255 variational inequality, 190 variational method, 10, 186, 193 visible light, 233 von Neumann detector, 147 wave function, 4 wave packet projection, 2 Weak interactions, 17 width, 50, 51, 273 Wigner–Weisskopf, 271 WKB approximation, 256 Yukawa potential, 12 Zeeman eﬀect, 74, 81 Zeeman interaction, 237, 238

The Ecole Polytechnique, one of France’s top academic institutions, has a longstanding tradition of producing exceptional scientiﬁc textbooks for its students. The original lecture notes, the Cours de l’Ecole Polytechnique, which were written by Cauchy and Jordan in the nineteenth century, are considered to be landmarks in the development of mathematics. The present series of textbooks is remarkable in that the texts incorporate the most recent scientiﬁc advances in courses designed to provide undergraduate students with the foundations of a scientiﬁc discipline. An outstanding level of quality is achieved in each of the seven scientiﬁc ﬁelds taught at the Ecole: pure and applied mathematics, mechanics, physics, chemistry, biology, and economics. The uniform level of excellence is the result of the unique selection of academic staff there which includes, in addition to the best researchers in its own renowned laboratories, a large number of world-famous scientists, appointed as part-time professors or associate professors, who work in the most advanced research centers France has in each ﬁeld. Another distinctive characteristic of these courses is their overall consistency; each course makes appropriate use of relevant concepts introduced in the other textbooks. This is because each student at the Ecole Polytechnique has to acquire basic knowledge in the seven scientiﬁc ﬁelds taught there, so a substantial link between departments is necessary. The distribution of these courses used to be restricted to the 900 students at the Ecole. Some years ago we were very successful in making these courses available to a larger French-reading audience. We now build on this success by making these textbooks also available in English.

Jean-Louis Basdevant Jean Dalibard

The Quantum Mechanics Solver How to Apply Quantum Theory to Modern Physics

Second Edition

With 59 Figures, Numerous Problems and Solutions

ABC

Professor Jean-Louis Basdevant

Professor Jean Dalibard

Department of Physics Laboratoire Leprince-Ringuet Ecole Polytechnique 91128 Palaisseau Cedex France E-mail: [email protected] polytechnique.edu

Ecole Normale Superieure Laboratoire Kastler Brossel rue Lhomond 24, 75231 Paris, CX 05 France E-mail: [email protected]

Library of Congress Control Number: 2005930228 ISBN-10 ISBN-13 ISBN-10 ISBN-13

3-540-27721-8 (2nd Edition) Springer Berlin Heidelberg New York 978-3-540-27721-7 (2nd Edition) Springer Berlin Heidelberg New York 3-540-63409-6 (1st Edition) Springer Berlin Heidelberg New York 978-3-540-63409-6 (1st Edition) Springer Berlin Heidelberg New York

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Preface to the Second Edition

Quantum mechanics is an endless source of new questions and fascinating observations. Examples can be found in fundamental physics and in applied physics, in mathematical questions as well as in the currently popular debates on the interpretation of quantum mechanics and its philosophical implications. Teaching quantum mechanics relies mostly on theoretical courses, which are illustrated by simple exercises often of a mathematical character. Reducing quantum physics to this type of problem is somewhat frustrating since very few, if any, experimental quantities are available to compare the results with. For a long time, however, from the 1950s to the 1970s, the only alternative to these basic exercises seemed to be restricted to questions originating from atomic and nuclear physics, which were transformed into exactly soluble problems and related to known higher transcendental functions. In the past ten or twenty years, things have changed radically. The development of high technologies is a good example. The one-dimensional squarewell potential used to be a rather academic exercise for beginners. The emergence of quantum dots and quantum wells in semiconductor technologies has changed things radically. Optronics and the associated developments in infrared semiconductor and laser technologies have considerably elevated the social rank of the square-well model. As a consequence, more and more emphasis is given to the physical aspects of the phenomena rather than to analytical or computational considerations. Many fundamental questions raised since the very beginnings of quantum theory have received experimental answers in recent years. A good example is the neutron interference experiments of the 1980s, which gave experimental answers to 50 year old questions related to the measurability of the phase of the wave function. Perhaps the most fundamental example is the experimental proof of the violation of Bell’s inequality, and the properties of entangled states, which have been established in decisive experiments since the late 1970s. More recently, the experiments carried out to quantitatively verify decoherence eﬀects and “Schr¨odinger-cat” situations have raised considerable

VI

Preface to the Second Edition

interest with respect to the foundations and the interpretation of quantum mechanics. This book consists of a series of problems concerning present-day experimental or theoretical questions on quantum mechanics. All of these problems are based on actual physical examples, even if sometimes the mathematical structure of the models under consideration is simpliﬁed intentionally in order to get hold of the physics more rapidly. ´ The problems have all been given to our students in the Ecole Polytech´ nique and in the Ecole Normale Sup´erieure in the past 15 years or so. A special ´ feature of the Ecole Polytechnique comes from a tradition which has been kept for more than two centuries, and which explains why it is necessary to devise original problems each year. The exams have a double purpose. On one hand, they are a means to test the knowledge and ability of the students. On the other hand, however, they are also taken into account as part of the entrance examinations to public oﬃce jobs in engineering, administrative and military careers. Therefore, the traditional character of stiﬀ competitive examinations and strict meritocracy forbids us to make use of problems which can be found in the existing literature. We must therefore seek them among the forefront of present research. This work, which we have done in collaboration with many colleagues, turned out to be an amazing source of discussions between us. We all actually learned very many things, by putting together our knowledge in our respective ﬁelds of interest. Compared to the ﬁrst version of this book, which was published by Springer-Verlag in 2000, we have made several modiﬁcations. First of all, we have included new themes, such as the progress in measuring neutrino oscillations, quantum boxes, the quantum thermometer etc. Secondly, it has appeared useful to include, at the beginning, a brief summary on the basics of quantum mechanics and the formalism we use. Finally, we have grouped the problems under three main themes. The ﬁrst (Part A) deals with Elementary Particles, Nuclei and Atoms, the second (Part B) with Quantum Entanglement and Measurement, and the third (Part C) with Complex Systems. We are indebted to many colleagues who either gave us driving ideas, or wrote ﬁrst drafts of some of the problems presented here. We want to pay a tribute to the memory of Gilbert Grynberg, who wrote the ﬁrst versions of “The hydrogen atom in crossed ﬁelds”, “Hidden variables and Bell’s inequalities” and “Spectroscopic measurement on a neutron beam”. We are particularly grateful to Fran¸cois Jacquet, Andr´e Roug´e and Jim Rich for illuminating discussions on “Neutrino oscillations”. Finally we thank Philippe Grangier, who actually conceived many problems among which the “Schr¨ odinger’s cat”, the “Ideal quantum measurement” and the “Quantum thermometer”, G´erald Bastard for “Quantum boxes”, Jean-No¨el Chazalviel for “Hyperﬁne structure in electron spin resonance”, Thierry Jolicoeur for “Magnetic excitons”, Bernard Equer for “Probing matter with positive muons”, Vincent Gillet for “Energy loss of ions in matter”, and Yvan Castin, Jean-Michel Courty and Do-

Preface to the Second Edition

VII

minique Delande for “Quantum reﬂection of atoms on a surface” and “Quantum motion in a periodic potential”. Palaiseau, April 2005

Jean-Louis Basdevant Jean Dalibard

Contents

Summary of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 General Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 The Particular Case of a Point-Like Particle; Wave Mechanics . 4 4 Angular Momentum and Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 5 Exactly Soluble Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 6 Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 7 Identical Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 8 Time-Evolution of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 9 Collision Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Part I Elementary Particles, Nuclei and Atoms 1

Neutrino Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Mechanism of the Oscillations; Reactor Neutrinos . . . . . . . . . . . 1.2 Oscillations of Three Species; Atmospheric Neutrinos . . . . . . . . 1.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 18 20 23 27

2

Atomic Clocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Hyperﬁne Splitting of the Ground State . . . . . . . . . . . . . . . . 2.2 The Atomic Fountain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The GPS System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 The Drift of Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . 2.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29 29 31 32 32 33

3

Neutron Interferometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Neutron Interferences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The Gravitational Eﬀect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Rotating a Spin 1/2 by 360 Degrees . . . . . . . . . . . . . . . . . . . . . . . .

37 38 39 40

X

Contents

3.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4

Spectroscopic Measurement on a Neutron Beam . . . . . . . . . . . 47 4.1 Ramsey Fringes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5

Analysis of a Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . 5.1 Preparation of the Neutron Beam . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Spin State of the Neutrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 The Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

Measuring the Electron Magnetic Moment Anomaly . . . . . . . 65 6.1 Spin and Momentum Precession of an Electron in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7

Decay of a Tritium Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The Energy Balance in Tritium Decay . . . . . . . . . . . . . . . . . . . . . 7.2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69 69 70 71

8

The Spectrum of Positronium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Positronium Orbital States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Hyperﬁne Splitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Zeeman Eﬀect in the Ground State . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Decay of Positronium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73 73 73 74 75 77

9

The Hydrogen Atom in Crossed Fields . . . . . . . . . . . . . . . . . . . . 9.1 The Hydrogen Atom in Crossed Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Pauli’s Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81

10 Energy Loss of Ions in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Energy Absorbed by One Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Energy Loss in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87 87 88 90 94

55 55 57 57 59

82 82 83

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XI

Part II Quantum Entanglement and Measurement 11 The EPR Problem and Bell’s Inequality . . . . . . . . . . . . . . . . . . . 99 11.1 The Electron Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 11.2 Correlations Between the Two Spins . . . . . . . . . . . . . . . . . . . . . . . 100 11.3 Correlations in the Singlet State . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 11.4 A Simple Hidden Variable Model . . . . . . . . . . . . . . . . . . . . . . . . . . 101 11.5 Bell’s Theorem and Experimental Results . . . . . . . . . . . . . . . . . . 102 11.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12 Schr¨ odinger’s Cat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12.1 The Quasi-Classical States of a Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 12.2 Construction of a Schr¨ odinger-Cat State . . . . . . . . . . . . . . . . . . . . 111 12.3 Quantum Superposition Versus Statistical Mixture . . . . . . . . . . . 111 12.4 The Fragility of a Quantum Superposition . . . . . . . . . . . . . . . . . . 112 12.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 12.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 13 Quantum Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 13.2 Correlated Pairs of Spins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 13.3 The Quantum Cryptography Procedure . . . . . . . . . . . . . . . . . . . . 125 13.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 14 Direct Observation of Field Quantization . . . . . . . . . . . . . . . . . . 131 14.1 Quantization of a Mode of the Electromagnetic Field . . . . . . . . 131 14.2 The Coupling of the Field with an Atom . . . . . . . . . . . . . . . . . . . 133 14.3 Interaction of the Atom with an “Empty” Cavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 14.4 Interaction of an Atom with a Quasi-Classical State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 14.5 Large Numbers of Photons: Damping and Revivals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 14.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 14.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 15 Ideal Quantum Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 15.1 Preliminaries: a von Neumann Detector . . . . . . . . . . . . . . . . . . . . 147 15.2 Phase States of the Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . 148 15.3 The Interaction between the System and the Detector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.4 An “Ideal” Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 15.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

XII

Contents

15.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 16 The Quantum Eraser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 16.1 Magnetic Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 16.2 Ramsey Fringes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 16.3 Detection of the Neutron Spin State . . . . . . . . . . . . . . . . . . . . . . . 158 16.4 A Quantum Eraser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 16.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 16.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 17 A Quantum Thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 17.1 The Penning Trap in Classical Mechanics . . . . . . . . . . . . . . . . . . . 169 17.2 The Penning Trap in Quantum Mechanics . . . . . . . . . . . . . . . . . . 170 17.3 Coupling of the Cyclotron and Axial Motions . . . . . . . . . . . . . . . 172 17.4 A Quantum Thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 17.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Part III Complex Systems 18 Exact Results for the Three-Body Problem . . . . . . . . . . . . . . . . 185 18.1 The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 18.2 The Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 18.3 Relating the Three-Body and Two-Body Sectors . . . . . . . . . . . . . 186 18.4 The Three-Body Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . 187 18.5 From Mesons to Baryons in the Quark Model . . . . . . . . . . . . . . . 187 18.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 19 Properties of a Bose–Einstein Condensate . . . . . . . . . . . . . . . . . 193 19.1 Particle in a Harmonic Trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 19.2 Interactions Between Two Conﬁned Particles . . . . . . . . . . . . . . . . 194 19.3 Energy of a Bose–Einstein Condensate . . . . . . . . . . . . . . . . . . . . . 195 19.4 Condensates with Repulsive Interactions . . . . . . . . . . . . . . . . . . . 195 19.5 Condensates with Attractive Interactions . . . . . . . . . . . . . . . . . . . 196 19.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 19.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 20 Magnetic Excitons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.1 The Molecule CsFeBr3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 20.2 Spin–Spin Interactions in a Chain of Molecules . . . . . . . . . . . . . . 204 20.3 Energy Levels of the Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 20.4 Vibrations of the Chain: Excitons . . . . . . . . . . . . . . . . . . . . . . . . . 206 20.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

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XIII

21 A Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 21.1 Results on the One-Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 21.2 The Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 21.3 Quantum Box in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . 218 21.4 Experimental Veriﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 21.5 Anisotropy of a Quantum Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 21.6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 21.7 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 22 Colored Molecular Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 22.1 Hydrocarbon Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 22.2 Nitrogenous Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 22.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 22.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 23 Hyperﬁne Structure in Electron Spin Resonance . . . . . . . . . . . 237 23.1 Hyperﬁne Interaction with One Nucleus . . . . . . . . . . . . . . . . . . . . 238 23.2 Hyperﬁne Structure with Several Nuclei . . . . . . . . . . . . . . . . . . . . 238 23.3 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 23.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 24 Probing Matter with Positive Muons . . . . . . . . . . . . . . . . . . . . . . 245 24.1 Muonium in Vacuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 24.2 Muonium in Silicon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 24.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 25 Quantum Reﬂection of Atoms from a Surface . . . . . . . . . . . . . . 255 25.1 The Hydrogen Atom–Liquid Helium Interaction . . . . . . . . . . . . . 255 25.2 Excitations on the Surface of Liquid Helium . . . . . . . . . . . . . . . . 257 25.3 Quantum Interaction Between H and Liquid He . . . . . . . . . . . . . 258 25.4 The Sticking Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 25.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 25.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 26 Laser Cooling and Trapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 26.1 Optical Bloch Equations for an Atom at Rest . . . . . . . . . . . . . . . 267 26.2 The Radiation Pressure Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 26.3 Doppler Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 26.4 The Dipole Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 26.5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 26.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

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27 Bloch Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 27.1 Unitary Transformation on a Quantum System . . . . . . . . . . . . . . 277 27.2 Band Structure in a Periodic Potential . . . . . . . . . . . . . . . . . . . . . 277 27.3 The Phenomenon of Bloch Oscillations . . . . . . . . . . . . . . . . . . . . . 278 27.4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 27.5 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Author Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

Summary of Quantum Mechanics

In the following pages we remind the basic deﬁnitions, notations and results of quantum mechanics.

1 Principles Hilbert Space The ﬁrst step in treating a quantum physical problem consists in identifying the appropriate Hilbert space to describe the system. A Hilbert space is a complex vector space, with a Hermitian scalar product. The vectors of the space are called kets and are noted |ψ. The scalar product of the ket |ψ1 and the ket |ψ2 is noted ψ2 |ψ1 . It is linear in |ψ1 and antilinear in |ψ2 and one has: ∗ ψ1 |ψ2 = (ψ2 |ψ1 ) . Deﬁnition of the State of a System; Pure Case The state of a physical system is completely deﬁned at any time t by a vector of the Hilbert space, normalized to 1, noted |ψ(t). Owing to the superposition principle, if |ψ1 and |ψ2 are two possible states of a given physical system, any linear combination |ψ ∝ c1 |ψ1 + c2 |ψ2 , where c1 and c2 are complex numbers, is a possible state of the system. These coeﬃcients must be chosen such that ψ|ψ = 1.

2

Summary of Quantum Mechanics

Measurement To a given physical quantity A one associates a self-adjoint (or Hermitian) operator Aˆ acting in the Hilbert space. In a measurement of the quantity A, ˆ the only possible results are the eigenvalues aα of A. Consider a system in a state |ψ. The probability P(aα ) to ﬁnd the result aα in a measurement of A is 2 P(aα ) = Pˆα |ψ , where Pˆα is the projector on the eigensubspace Eα associated to the eigenvalue aα . After a measurement of Aˆ which has given the result aα , the state of the system is proportional to Pˆα |ψ (wave packet projection or reduction). A single measurement gives information on the state of the system after the measurement has been performed. The information acquired on the state before the measurement is very “poor”, i.e. if the measurement gave the result aα , one can only infer that the state |ψ was not in the subspace orthogonal to Eα . In order to acquire accurate information on the state before measurement, one must use N independent systems, all of which are prepared in the same state |ψ (with N 1) . If we perform N1 measurements of Aˆ1 (eigenvalˆ ues p {a1,α }), N2 measurements of A2 (eigenvalues {a2,α }), and so on (with i=1 Ni = N ), we can determine the probability distribution of the ai,α , and therefore the Pˆi,α |ψ 2 . If the p operators Aˆi are well chosen, this determines unambiguously the initial state |ψ. Evolution When the system is not being measured, the evolution of its state vector is given by the Schr¨odinger equation i¯ h

d ˆ |ψ = H(t) |ψ(t) , dt

ˆ where the hermitian operator H(t) is the Hamiltonian, or energy observable, of the system at time t. If we consider an isolated system, whose Hamiltonian is time-independent, the energy eigenstates of the Hamiltonian |φn are the solution of the time independent Schr¨ odinger equation: ˆ n = En |φn . H|φ They form an orthogonal basis of the Hilbert space. This basis is particularly useful. If we decompose the initial state |ψ(0) on this basis, we can immediately write its expression at any time as:

1 Principles

|ψ(0) =

αn |φn

→

|ψ(t) =

n

3

αn e−iEn t/¯h |φn .

n

The coeﬃcients are αn = φn |ψ(0), i.e. e−iEn t/¯h |φn φn |ψ(0) . |ψ(t) = n

Complete Set of Commuting Observables (CSCO) ˆ B, ˆ . . . , X} ˆ is a CSCO if all of these operators commute A set of operators {A, and if their common eigenbasis {|α, β, . . . , ξ} is unique (up to a phase factor). In that case, after the measurement of the physical quantities {A, B, . . . , X}, the state of the system is known unambiguously. If the measurements have ˆ the state of the system is given the values α for A, β for B, . . . , ξ for X, |α, β, . . . , ξ. Entangled States Consider a quantum system S formed by two subsystems S1 and S2 . The Hilbert space in which we describe S is the tensor product of the Hilbert spaces E1 and E2 respectively associated with S1 and S2 . If we note {|αm } a basis of S1 and {|βn } a basis of S2 , a possible basis of the global system is {|αm ⊗ |βn }. Any state vector of the global system can be written as: Cm,n |αm ⊗ |βn . |Ψ = m,n

If this vector can be written as |Ψ = |α ⊗ |β, where |α and |β are vectors of E1 and E2 respectively, one calls it a factorized state. In general an arbitrary state |Ψ is not factorized: there are quantum correlations between the two subsystems, and |Ψ is called an Entangled state. Statistical Mixture and the Density Operator If we have an incomplete information on the state of the system, for instance because the measurements are incomplete, one does not know exactly its state vector. The state can be described by a density operator ρˆ whose properties are the following: • The density operator is hermitian and its trace is equal to 1. • All the eigenvalues Πn of the density operator are non-negative. The density operator can therefore be written as Πn |φn φn | , ρˆ = n

4

Summary of Quantum Mechanics

where the |φn are the eigenstates of ρˆ and the Πn can be interpreted as a probability distribution. In the case of a pure state, all eigenvalues vanish except one which is equal to 1. • The probability to ﬁnd the result aα in a measurement of the physical quantity A is given by ˆ n . Πn φn |A|φ P(aα ) = Tr Pˆα ρˆ = n

The state of the system after the measurement is ρˆ ∝ Pˆα ρˆPˆα . • As long as the system is not measured, the evolution of the density operator is given by d ˆ , ρˆ(t)] . i¯ h ρˆ(t) = [H(t) dt

2 General Results Uncertainty Relations Consider 2N physical systems which are identical and independent, and are all prepared in the same state |ψ (we assume N 1). For N of them, we measure a physical quantity A, and for the N others , we measure a physical quantity B. The rms deviations ∆a and ∆b of the two series of measurements satisfy the inequality 1 ˆ B]|ψ ˆ ∆a ∆b ≥ ψ|[A, . 2 Ehrenfest Theorem ˆ Consider a system which evolves under the action of a Hamiltonian H(t), and ˆ an observable A(t). The expectation value of this observable evolves according to the equation: d 1 ∂ Aˆ ˆ H]|ψ ˆ a = ψ|[A, + ψ| |ψ . dt i¯ h ∂t ˆ the expecIn particular, if Aˆ is time-independent and if it commutes with H, tation value a is a constant of the motion.

3 The Particular Case of a Point-Like Particle; Wave Mechanics The Wave Function For a point-like particle for which we can neglect possible internal degrees of freedom, the Hilbert space is the space of square integrable functions (written in mathematics as L2 (R3 )).

3 The Particular Case of a Point-Like Particle; Wave Mechanics

5

The state vector |ψ is represented by a wave function ψ(r). The quantity |ψ(r)|2 is the probability density to ﬁnd the particle at point r in dimensional space. Its Fourier transform ϕ(p): 1 ϕ(p) = e−ip·r/¯h ψ(r) d3 r (2π¯ h)3/2 is the probability amplitude to ﬁnd that the particle has a momentum p. Operators Among the operators associated to usual physical quantities, one ﬁnds: • The position operator rˆ ≡ (ˆ x, yˆ, zˆ), which consists in multiplying the wave function ψ(r) by r. ˆ whose action on the wave function ψ(r) is the • The momentum operator p operation −i¯ h∇. • The Hamiltonian, or energy operator, for a particle placed in a potential V (r): 2 ˆ = pˆ + V (ˆ r) H 2M

→

¯2 2 h ˆ Hψ(r) =− ∇ ψ(r) + V (r)ψ(r) , 2M

where M is the mass of the particle. Continuity of the Wave Function If the potential V is continuous, the eigenfunctions of the Hamiltonian ψα (r) are continuous and so are their derivatives. This remains true if V (r) is a step function: ψ and ψ are continuous where V (r) has discontinuities. In the case of inﬁnitely high potential steps, (for instance V (x) = +∞ for x < 0 and V (x) = 0 for x ≥ 0), ψ(x) is continuous and vanishes at the discontinuity of V (ψ(0) = 0), while its ﬁrst derivative ψ (x) is discontinuous. In one dimension, it is interesting to consider potentials which are Dirac distributions, V (x) = g δ(x). The wave function is continuous and the discontinuity of its derivative is obtained by integrating the Schr¨ odinger equation around the center of the delta function [ψ (0+ ) − ψ (0− ) = (2M g/¯h2 ) ψ(0) in our example]. Position-Momentum Uncertainty Relations Using the above general result, one ﬁnds: [ˆ x, pˆx ] = i¯ h

→

∆x ∆px ≥ ¯h/2 ,

and similar relations for the y and z components.

6

Summary of Quantum Mechanics

4 Angular Momentum and Spin Angular Momentum Observable An angular momentum observable Jˆ is a set of three operators {Jˆx , Jˆy , Jˆz } which satisfy the commutation relations h Jˆz , [Jˆx , Jˆy ] = i¯

[Jˆy , Jˆz ] = i¯ h Jˆx ,

[Jˆz , Jˆx ] = i¯h Jˆy .

ˆ = rˆ × p ˆ is an The orbital angular momentum with respect to the origin L angular momentum observable. The observable Jˆ2 = Jˆx2 + Jˆy2 + Jˆz2 commutes with all the components ˆ Ji . One can therefore ﬁnd a common eigenbasis of Jˆ2 and one of the three components Jˆi . Traditionally, one chooses i = z. Eigenvalues of the Angular Momentum The eigenvalues of Jˆ2 are of the form h ¯ 2 j(j + 1) with j integer or half integer. 2 ˆ In an eigensubspace of J corresponding to a given value of j, the eigenvalues of Jˆz are of the form hm , ¯

with m ∈ {−j, −j + 1, . . . , j − 1, j}

(2j + 1 values) .

The corresponding eigenstates are noted |α, j, m, where α represents the other quantum numbers which are necessary in order to deﬁne the states completely. The states |α, j, m are related to |α, j, m ± 1 by the operators Jˆ± = Jˆx ± iJˆy : Jˆ± |α, j, m = j(j + 1) − m(m ± 1) |α, j, m ± 1 . Orbital Angular Momentum of a Particle In the case of an orbital angular momentum, only integer values of j and m are allowed. Traditionally, one notes j = in this case. The common eigenstates ˆ z can be written in spherical coordinates as R(r) Y,m (θ, ϕ), ˆ 2 and L ψ(r) of L where the radial wave function R(r) is arbitrary and where the functions Y,m are the spherical harmonics, i.e. the harmonic functions on the sphere of radius one. The ﬁrst are: 3 1 cos θ , , Y1,0 (θ, ϕ) = Y0,0 (θ, ϕ) = √ 4π 4π 3 3 iϕ sin θ e , Y1,−1 (θ, ϕ) = sin θ e−iϕ . Y1,1 (θ, ϕ) = − 8π 8π

5 Exactly Soluble Problems

7

Spin In addition to its angular momentum, a particle can have an intrinsic angular momentum called its Spin. The spin, which is noted traditionally j = s, can take half-integer as well as integer values. The electron, the proton, the neutron are spin s = 1/2 particles, for which the projection of the intrinsic angular momentum can take either of the two values m¯h: m = ±1/2. In the basis |s = 1/2 , m = ±1/2, the operators Sˆx , Sˆy , Sˆz have the matrix representations:

h 01 ¯ h 0 −i ¯ ¯h 1 0 Sˆx = , Sˆy = , Sˆz = . 2 10 2 i 0 2 0 −1 Addition of Angular Momenta Consider a system S made of two subsystems S1 and S2 , of angular momenta ˆ 2 . The observable Jˆ = Jˆ1 + Jˆ2 is an angular momentum observable. ˆ 1 and J J In the subspace corresponding to given values j1 and j2 (of dimension (2j1 + 1) × (2j2 + 1)), the possible values for the quantum number j corresponding to the total angular momentum of the system Jˆ are: j = |j1 − j2 | , |j1 − j2 | + 1 , · · · , j1 + j2 , with, for each value of j, the 2j + 1 values of m: m = −j, −j + 1, · · · , j. For instance, adding two spins 1/2, one can obtain an angular momentum 0 (singlet state j = m = 0) and three states of angular momentum 1 (triplet states j = 1, m = 0, ±1). The relation between the factorized basis |j1 , m1 ⊗ |j2 , m2 and the total angular momentum basis |j1 , j2 ; j, m is given by the Clebsch-Gordan coeﬃcients: j,m Cj1 ,m1 ;j2 ,m2 |j1 , m1 ⊗ |j2 , m2 . |j1 , j2 ; j, m = m1 m2

5 Exactly Soluble Problems The Harmonic Oscillator For simplicity, we consider the one-dimensional problem. The harmonic potential is written V (x) = mω 2 x2 /2. The natural length and momentum scales are √ h ¯ , p0 = ¯hmω . x0 = mω ˆ =x By introducing the reduced operators X ˆ/x0 and Pˆ = pˆ/p0 , the Hamiltonian is:

8

Summary of Quantum Mechanics

¯ ω ˆ2 ˆ =h ˆ2 , H P +X 2

ˆ Pˆ ] = i . with [X,

We deﬁne the creation and annihilation operators a ˆ† and a ˆ by: 1 ˆ 1 ˆ + iPˆ , a ˆ† = √ X − iPˆ , [ˆ a, a ˆ† ] = 1 . a ˆ= √ X 2 2 One has

†

ˆ =h H ¯ω a ˆ a ˆ + 1/2 .

ˆ are (n + 1/2)¯ The eigenvalues of H hω, with n non-negative integer. These eigenvalues are non-degenerate. The corresponding eigenvectors are noted |n. We have: √ a ˆ† |n = n + 1 |n + 1 and √ a ˆ|n = n |n − 1 if n > 0 , = 0 if n = 0 . The corresponding wave functions are the Hermite functions. The ground state |n = 0 is given by: ψ0 (x) =

1 √ exp(−x2 / 2x20 ) . π 1/4 x0

Higher dimension harmonic oscillator problems are deduced directly from these results. The Coulomb Potential (bound states) We consider the motion of an electron in the electrostatic ﬁeld of the proton. We note µ the reduced mass (µ = me mp /(me + mp ) me ) and we set e2 = q 2 /(4π0 ). Since the Coulomb potential is rotation invariant, we can ﬁnd ˆ z . The bound ˆ to L ˆ 2 and to L a basis of states common to the Hamiltonian H, states are characterized by the 3 quantum numbers n, , m with: ψn,,m (r) = Rn, (r) Y,m (θ, ϕ) , where the Y,m are the spherical harmonics. The energy levels are of the form En = −

EI n2

with EI =

µe4 13.6 eV . 2¯ h2

The principal quantum number n is a positive integer and can take all integer values from 0 to n − 1. The total degeneracy (in m and ) of a given energy level is n2 (we do not take spin into account). The radial wave functions Rn, are of the form:

6 Approximation Methods

Rn, (r) = r Pn, (r) exp(−r/(na1 )) ,

with

a1 =

9

¯2 h 0.53 ˚ A. µe2

Pn, (r) is a polynomial of degree n − − 1 called a Laguerre polynomial. The length a 1 is the Bohr radius. The ground state wave function is ψ1,0,0 (r) = e−r/a1 / πa31 .

6 Approximation Methods Time-Independent Perturbations ˆ which can be written as H ˆ = We consider a time-independent Hamiltonian H ˆ 0 + λH ˆ 1 . We suppose that the eigenstates of H ˆ 0 are known: H ˆ 0 |n, r = En |n, r , H

r = 1, 2, . . . , pn

ˆ 1 is where pn is the degeneracy of En . We also suppose that the term λH suﬃciently small so that it only results in small perturbations of the spectrum ˆ 0. of H ˆ which Non-degenerate Case. In this case, pn = 1 and the eigenvalue of H coincides with En as λ → 0 is given by: ˜n = En + λ n|H ˆ 1 |n + λ2 E

|k|H ˆ 1 |n|2 + O(λ3 ) . En − E k

k=n

The corresponding eigenstate is: |ψn = |n + λ

k|H ˆ 1 |n |k + O(λ2 ) En − E k

k=n

ˆ at ﬁrst order in Degenerate Case. In order to obtain the eigenvalues of H λ, and the corresponding eigenstates, one must diagonalize the restriction of ˆ 0 associated with the eigenvalue En , i.e. ﬁnd the pn ˆ 1 to the subspace of H λH solutions of the “secular” equation: ˆ 1 |n, 1 − ∆E ˆ 1 |n, pn n, 1|λH ... n, 1|λH .. .. = 0. ˆ . . n, r|λH1 |n, r − ∆E n, p |λH ˆ ˆ ... n, pn |λH1 |n, pn − ∆E n 1 |n, 1 ˜n,r = En +∆Er , r = 1, . . . , pn . In general, The energies to ﬁrst order in λ are E the perturbation is lifted (at least partially) by the perturbation.

10

Summary of Quantum Mechanics

Variational Method for the Ground State Consider an arbitrary state |ψ normalized to 1. The expectation value of the energy in this state is greater than or equal to the ground state energy E0 : ˆ ψ|H|ψ ≥ E0 . In order to ﬁnd an upper bound to E0 , one uses a set of trial wave functions which depend on a set of parameters, and one looks for the minimum of E for these functions. This minimum always lies above E0 .

7 Identical Particles All particles in nature belong to one of the following classes: • Bosons, which have integer spin. The state vector of N identical bosons is totally symmetric with respect to the exchange of any two of these particles. • Fermions, which have half-integer spin. The state vector of N identical fermions is totally antisymmetric with respect to the exchange of any two of these particles. Consider a basis {|ni , i = 1, 2, . . .} of the one particle Hilbert space. Consider a system of N identical particles, which we number arbitrarily from 1 to N . (a) If the particles are bosons, the state vector of the system with N1 particles in the state |n1 , N2 particles in the state |n2 , etc., is: 1 1 |1 : nP (1) ; 2 : nP (2) ; . . . ; N : nP (N ) , |Ψ = √ √ N ! N1 !N2 ! · · · P where the summation is made on the N ! permutations of a set of N elements. (b) If the particles are fermions, the state corresponding to one particle in the state |n1 , another in the state |n2 , etc., is given by the Slater determinant: |1 : n1 |1 : n2 . . . |1 : nN 1 |2 : n1 |2 : n2 . . . |2 : nN |Ψ = √ . .. .. .. N ! . . . |N : n1 |N : n2 . . . |N : nN Since the state vector is antisymmetric, two fermions cannot be in the same quantum state (Pauli’s exclusion principle). The above states form a basis of the N −fermion Hilbert space.

8 Time-Evolution of Systems

11

8 Time-Evolution of Systems Rabi Oscillation ˆ0 = h ¯ ω0 |++|. We couple Consider a two-level system |±, of Hamiltonian H ˆ these two states with a Hamiltonian H1 :

hω1 −iωt ˆ1 = ¯ e H |+−| + eiωt |−+| . 2 We assume that the state of the system is |− at time t = 0. The probability to ﬁnd the system in the state |+ at time t is: P (t) =

ω12 sin2 (ΩT /2) Ω2

with Ω 2 = (ω − ω0 )2 + ω12 .

Time-Dependent Perturbation Theory ˆ 1 (t). We assume ˆ ˆ0 + H We consider a system whose Hamiltonian is H(t) =H ˆ 0 and the corresponding energies En are known. At the eigenstates |n of H ˆ 0 . To ﬁrst time t = 0, we assume that the system is in the eigenstate |i of H ˆ 1 , the probability amplitude to ﬁnd the system in another eigenstate order in H |f at time t is: 1 t i(Ef −Ei )t/¯h ˆ 1 (t )|i dt . a(t) = e f |H i¯ h 0 In the case of a time-independent perturbation H1 , the probability is: P (t) = |a(t)|2 =

2 2 1 ˆ 1 |i sin (ωt/2) , | H f (ω/2)2 h2 ¯

where we have set ¯hω = Ef − Ei . Fermi’s Golden Rule and Exponential Decay ˆ 0 . Initially, the system Consider a system with an unperturbed Hamiltonian H is in an eigenstate |i of energy Ei . We assume that this system is coupled to ˆ 0 by the time-independent perturbation a continuum {|f } of eigenstates of H Vˆ . For simplicity, we assume that the matrix elements f |Vˆ |i only depend on the energies Ef of the states |f . To lowest order in Vˆ , this coupling results in a ﬁnite lifetime τ of the state |i: the probability to ﬁnd the system in the state |i at time t > 0 is e−t/τ with: 2π 1 = |f |Vˆ |i|2 ρ(Ei ) . τ h ¯ The matrix element f |Vˆ |i is evaluated for a state |f of energy Ef = Ei . The function ρ(E) is the density of ﬁnal states. For non relativistic particles

12

Summary of Quantum Mechanics

(E = p2 /2m) or ultra-relativistic particles (E = cp, for instance photons), its values are respectively: √ mL3 2mE L3 E 2 ρ (E) = . ρnon rel. (E) = ultra rel. 2π 2 ¯ h3 2π 2 ¯h3 c3 When the spin degree of freedom of the particle comes into play, this density of state must be multiplied by the number of possible spin states 2s + 1, where s is the spin of the particle. The quantity L3 represents the normalization volume (and cancels identically with the normalization factors of the states |i and |f ). Consider an atomic transition treated as a two-level system, an excited state |e and a ground state |g, separated by an energy h ¯ ω and coupled via an electric dipole interaction. The lifetime τ of the excited state due to this spontaneous emission is given by: ω3 1 ˆ 2 e| D|g = , τ 3π0 ¯ hc3 ˆ is the electric dipole operator. where D

9 Collision Processes Born Approximation We consider an elastic collision process of a non-relativistic particle of mass m with a ﬁxed potential V (r). To second order in V , the elastic scattering cross-section for an incident particle in the initial momentum state p and the ﬁnal momentum state p is given by: dσ = dΩ

m 2π¯ h2

2

|V˜ (p − p )|2 ,

with V˜ (q) =

eiq·r/¯h V (r) d3 r .

Example: the Yukawa potential. We consider V (r) = g

¯ c −r/a h e , r

which gives, writing p = h ¯ k: dσ = dΩ

2mgca2 h ¯

2

1 1+

4a2 k 2

2 sin2 (θ/2)

(Born) ,

where θ is the scattering angle between p and p . The total cross-section is then:

2 4πa2 2 mgca (Born) . σ(k) = h ¯ 1 + 4k 2 a2

9 Collision Processes

13

In the case where the range a of the potential tends to inﬁnity, we recover the Coulomb cross section:

2 g¯ hc 1 dσ = (exact) , dΩ 4E sin4 (θ/2) where E = p2 /(2m). Scattering by a Bound State We consider a particle a of mass m undergoing an elastic scattering on a system composed of n particles b1 , . . . , bn . These n particles form a bound state whose wave function is ψ0 (r 1 , . . . , r n ). In Born approximation, the cross section is

2 m dσ V˜j (q) Fj (q) . = |V(p − p )|2 with V(q) = 2 dΩ 2π¯ h j The potential Vj represents the interaction between particles a and bj . The form factor Fj is deﬁned by: Fj (q) = eiq·rj /¯h |ψ0 (r 1 , ..., r j , ..., r n )|2 d3 r1 . . . d3 rj . . . d3 rn . In general, interference eﬀects can be observed between the various q contributing to the sum which deﬁnes V(q). In the case of a charge distribution, V˜ is the Rutherford amplitude, and the form factor F is the Fourier transform of the charge density. General Scattering Theory In order to study the general problem of the scattering of a particle of mass m by a potential V (r), it is useful to determine the positive energy E = ˆ = pˆ2 /(2m) + V (r) whose asymptotic form is h2 k 2 /(2m) eigenstates of H ¯ ψk (r)

∼

|r|→∞

eik·r + f (k, u, u )

eikr . r

This corresponds to the superposition of an incident plane wave eik·r and a scattered wave. Such a state is called a stationary scattering state. The scattering amplitude f depends on the energy, on the incident direction u = k/k, and on the ﬁnal direction u = r/r. The diﬀerential cross section is given by: dσ = |f (k, u, u )|2 . dΩ The scattering amplitude is given by the implicit equation m e−ik ·r V (r ) ψk (r ) d3 r with k = ku . f (k, u, u ) = − 2 2π¯ h We recover Born’s approximation by choosing ψk (r ) eik·r .

14

Summary of Quantum Mechanics

Low Energy Scattering When the wavelength of the incident particle λ ∼ k −1 is large compared to the range of the potential, the amplitude f does not depend on u and u (at least if the potential decreases faster than r−3 at inﬁnity). The scattering is isotropic. The limit as = − limk→0 f (k) is called the scattering length.

Part I

Elementary Particles, Nuclei and Atoms

2 Atomic Clocks

We are interested in the ground state of the external electron of an alkali atom (rubidium, cesium,...). The atomic nucleus has a spin sn (sn = 3/2 for 87 Rb, sn = 7/2 for 133 Cs), which carries a magnetic moment µn . As in the case of atomic hydrogen, the ground state is split by the hyperﬁne interaction between the electron magnetic moment and the nuclear magnetic moment µn . This splitting of the ground state is used to devise atomic clocks of high accuracy, which have numerous applications such as ﬂight control in aircrafts, the G.P.S. system, the measurement of physical constants etc. In all the chapter, we shall neglect the eﬀects due to internal core electrons.

2.1 The Hyperﬁne Splitting of the Ground State 2.1.1. Give the degeneracy of the ground state if one neglects the magnetic interaction between the nucleus and the external electron. We note |me ; mn = |electron: se = 1/2, me ⊗ |nucleus: sn , mn a basis of the total spin states (external electron + nucleus). 2.1.2. We now take into account the interaction between the electron magnetic moment µe and the nuclear magnetic moment µn . As in the hydrogen atom, one can write the corresponding Hamiltonian (restricted to the spin subspace) as: ˆn , ˆe · S ˆ = A S H h2 ¯ ˆ e and S ˆ n are the spin operators where A is a characteristic energy, and where S of the electron and the nucleus, respectively. We want to ﬁnd the eigenvalues of this Hamiltonian. We introduce the operators Sˆe,± = Sˆe,x ± iSˆe,y and Sˆn,± = Sˆn,x ± iSˆn,y .

30

2 Atomic Clocks

(a) Show that ˆ = A Sˆe,+ Sˆn,− + Sˆe,− Sˆn,+ + 2Sˆe,z Sˆn,z . H 2 2¯ h (b) Show that the two states |me = 1/2; mn = sn

and

|me = −1/2; mn = −sn

ˆ and give the corresponding eigenvalues. are eigenstates of H, ˆ on the state |me = 1/2; mn with mn = sn ? (c) What is the action of H ˆ What is the action of H on the state |me = −1/2; mn with mn = −sn ? ˆ can be calculated (d) Deduce from these results that the eigenvalues of H by diagonalizing 2 × 2 matrices of the type:

A mn sn (sn + 1) − mn (mn + 1) . sn (sn + 1) − mn (mn + 1) −(mn + 1) 2 ˆ splits the ground state in two substates of energies E1 = 2.1.3. Show that H E0 + Asn /2 and E2 = E0 − A(1 + sn )/2. Recover the particular case of the hydrogen atom. 2.1.4. What are the degeneracies of the two sublevels E1 and E2 ? 2.1.5. Show that the states of energies E1 and E2 are eigenstates of the 2 ˆe + S ˆ n . Give the corresponding value s of ˆ2 = S square of the total spin S the spin.

H=1 m

Electromagnetic cavity Cold atoms Fig. 2.1. Sketch of the principle of an atomic clock with an atomic fountain, using laser-cooled atoms

2.2 The Atomic Fountain

31

2.2 The Atomic Fountain The atoms are initially prepared in the energy state E1 , and are sent upwards (Fig. 2.1). When they go up and down they cross a cavity where an electromagnetic wave of frequency ω is injected. This frequency is close to h. At the end of the descent, one detects the number of ω0 = (E1 − E2 )/¯ atoms which have ﬂipped from the E1 level to the E2 level. In all what follows, the motion of the atoms in space (free fall) is treated classically. It is only the evolution of their internal state which is treated quantum-mechanically. In order to simplify things, we consider only one atom in the sub-level of energy E1 . This state (noted |1) is coupled by the electromagnetic wave to only one state (noted |2) of the sublevel of energy E2 . By convention, we ﬁx ¯ ω0 /2, E2 = −¯ hω0 /2. We the origin of energies at (E1 + E2 )/2, i.e. E1 = h assume that the time to cross the cavity is very brief and that this crossing results in an evolution of the state vector of the form: |ψ(t) = α|1 + β|2 −→ |ψ(t + ) = α |1 + β |2 ,

1 α α 1 −ie−iωt √ with: . = iωt β 1 β 2 −ie 2.2.1. The initial state of the atom is |ψ(0) = |1. We consider a single round-trip of duration T , during which the atom crosses the cavity between t = 0 and t = , then evolves freely during a time T − 2, and crosses the cavity a second time between T − and T . Taking the limit → 0, show that the state of the atom after this round-trip is given by: |ψ(T ) = i e−iωT /2 sin((ω − ω0 )T /2) |1 − i eiωT /2 cos((ω − ω0 )T /2) |2 (2.1) 2.2.2. Give the probability P (ω) to ﬁnd an atom in the state |2 at time T . Determine the half-width ∆ω of P (ω) at the resonance ω = ω0 . What is the values of ∆ω for a 1 meter high fountain? We recall the acceleration of gravity g = 9.81 ms−2 . 2.2.3. We send a pulse of N atoms (N 1). After the round-trip, each atom is in the state given by (2.1). We measure separately the numbers of atoms in the states |1 and |2, which we note N1 and N2 (with N1 + N2 = N ). What is the statistical distribution of the random variables N1 and N2 ? Give their mean values and their r.m.s. deviations ∆Ni . Set φ = (ω − ω0 )T /2 and express the results in terms of cos φ, sin φ and N . 2.2.4. The departure from resonance |ω − ω0 | is characterized by the value of cos((ω − ω0 )T ) = N2 − N1 /N . Justify this formula. Evaluate the uncertainty ∆|ω − ω0 | introduced by the random nature of the variable N2 − N1 . Show that this uncertainty depends on N , but not on φ. 2.2.5. In Fig. 2.2 we have represented the precision of an atomic clock as a function of the number N of atoms per pulse. Does this variation with N agree with the previous results?

32

2 Atomic Clocks

Relative precision 10-13

N 10-14 104

105

106

Fig. 2.2. Relative accuracy ∆ω/ω of a fountain atomic clock as a function of the number of atoms N sent in each pulse

2.3 The GPS System The GPS system uses 24 satellites orbiting around the Earth at 20 000 km. Each of them contains an atomic clock. Each satellite sends, at equal spaced time intervals, an electromagnetic signal composed of a “click” from a clock and the indication of its position. A reception device on Earth, which does not have an atomic clock, detects the signals coming from several satellites. With its own (quartz) clock, it compares the times at which diﬀerent “clicks” arrive. 2.3.1. What is the minimum number of satellites that one must see at a given time in order to be able to position oneself in latitude, in longitude, and in altitude on the surface of the Earth? 2.3.2. We assume that the relative accuracy of each clock is ∆ω/ω = 10−13 and that the clocks are synchronized every 24 hours. What is the order of magnitude of the accuracy of the positioning just before the clocks undergo a new synchronization?

2.4 The Drift of Fundamental Constants Some cosmological models predict a (small) variation in time of the ﬁne struchc) ∼ 1/137. In order to test such an assumption, one ture constant α = e2 /(¯ can compare two atomic clocks, one using rubidium (Z = 37) atoms, the other cesium (Z = 55) atoms. In fact, one can show that the hyperﬁne splitting of an alkali atom varies approximately as:

11 2 2 ¯ ω0 ∝ α 1 + (αZ) for (αZ)2 1 . E1 − E2 = h 6

2.5 Solutions

33

By comparing a rubidium and a cesium clock at a one year interval, no sig(Cs) (Rb) was observed. More precisely, niﬁcant variation of the ratio R = ω0 /ω0 the relative variation |δR|/R is smaller than the experimental uncertainty, estimated to be 3 × 10−15 . What upper bound can one set on the relative variation rate |α/α|? ˙

2.5 Solutions Section 2.1: Hyperﬁne Splitting of the Ground State 2.1.1. The Hilbert space of the ground state is the tensor product of the electron spin space and the nucleus spin space. Its dimension d is therefore the product of their dimensions, i.e. d = 2 × (2sn + 1). 2.1.2. Energy levels of the hyperﬁne Hamiltonian. (a) Making use of 1 ˆ i ˆ Se,+ + Sˆe,− , Sˆe,y = Se,− − Sˆe,+ , Sˆe,x = 2 2 and a similar relation for Sˆn,x and Sˆn,y , one obtains the wanted result. (b) The action of Sˆe,+ Sˆn,− and Sˆe,− Sˆn,+ on |me = 1/2; mn = sn gives the null vector. The same holds for |me = −1/2; mn = −sn . Therefore, only the term contributes Sˆe,z Sˆn,z and one ﬁnds: ˆ |me = 1/2; mn = sn = Asn |me = 1/2; mn = sn H 2 ˆ |me = −1/2; mn = −sn = Asn |me = −1/2; mn = −sn . H 2 (c) We ﬁnd: Amn ˆ H|1/2; mn = |1/2; mn 2 A + sn (sn + 1) − mn (mn + 1) | − 1/2; mn + 1 2 ˆ − 1/2; mn = − Amn | − 1/2; mn H| 2 A + sn (sn + 1) − mn (mn − 1) |1/2; mn − 1 . 2 (d) >From the previous question, one concludes that the 2-dimensional subspaces Emn generated by |1/2; mn and | − 1/2; mn + 1 are globally stable ˆ The determination of the eigenvalues of H ˆ therefore under the action of H. consists in diagonalizing the series of 2 × 2 matrices corresponding to its reˆ striction to these subspaces. The matrix corresponding to the restriction of H to the subspace Emn is the same as given in the text.

34

2 Atomic Clocks

2.1.3. The eigenvalues given in the text are actually independent of mn . They are Asn /2 and −A(1 + sn )/2. In the case sn = 1/2 (hydrogen atom), these two eigenvalues are A/4 and −3A/4. 2.1.4. There are 2sn 2 × 2 matrices to be diagonalized, each of which gives a vector associated to Asn /2 and a vector associated to −A(1 + sn )/2. In addition we have found two independent eigenvectors, |1/2, sn and | − 1/2, −sn , associated to the eigenvalue Asn /2. We therefore obtain: Asn /2

degenerated 2sn + 2 times

−A(1 + sn /2)

degenerated 2sn times

We do recover the dimension 2(2sn + 1) of the total spin space of the ground state. 2.1.5. The square of the total spin is: 2 ˆ2 + S ˆ 2 + 2S ˆe · S ˆn = S ˆ2 + S ˆ 2 + 2¯h H ˆ2 = S ˆ . S e n e n A

ˆ 2 are proportional to the identity and are respectively: ˆ 2 and S The operators S e n h2 ˆ 2 = 3¯ S e 4

ˆ2 = h ¯ 2 sn (sn + 1) . S n 2

ˆ . More precisely, an eigenˆ is therefore an eigenstate of S An eigenstate of H ˆ 2 with eigenvalue ˆ state of H with eigenvalue Asn /2 is an eigenstate of S h2 (sn + 1/2)(sn + 3/2), corresponding to a total spin s = sn + 1/2. An eigen¯ ˆ 2 with eigenvalue ˆ with eigenvalue −A(1+sn )/2 is an eigenstate of S state of H h2 (sn − 1/2)(sn + 1/2), i.e. a total spin s = sn − 1/2. ¯ Section 2.2: The Atomic Fountain 2.2.1. In the limit → 0, the ﬁnal state vector of the atom is simply the matrix product: −iω T /2

1 e 0 0 α 1 −ie−iωT × = β 1 0 eiω0 T /2 2 −ieiωT

1 −i 1 × , −i 1 0 which corresponds to crossing the cavity, at time t = 0, then to a free evolution between t = 0 and t = T , then a second crossing of the cavity at time t = T . We therefore obtain the state vector of the text. 2.2.2. One ﬁnds P (ω) = |β |2 = cos2 ((ω − ω0 )T /2). This probability is equal π/(2T ). to 1 if one sits exactly at the resonance (ω = ω0 ). It is 1/2 if ω = ω0 ± For a round-trip free fall motion of height H = 1 m, we have T = 2 2H/g, i.e. T = 0.9 s and ∆ω = 1.7 s−1 .

2.5 Solutions

35

2.2.3. The detection of each atom gives the result E1 with a probability sin2 φ and E2 with a probability cos2 φ. Since the atoms are assumed to be independent, the distributions of the random variables N1 and N2 are binomial laws. We therefore have: √ N2 = N cos2 φ ∆N1 = ∆N2 = N | cos φ sin φ| . N1 = N sin2 φ 2.2.4. We do obtain N2 −N1 /N = cos 2φ = cos((ω −ω0 )T ). The ﬂuctuation on the variable N2 − N1 induces a ﬂuctuation on the determination of ω − ω0 . The two ﬂuctuations are related by: ∆(N2 − N1 ) = 2 |sin(2φ)| ∆φ . N √ √ Since ∆(N2 − N1 ) = 2 ∆N2 = N |sin 2φ|, we deduce ∆φ = 1/(2 N ), or equivalently: 1 √ . ∆|ω − ω0 | = 2T N The longer the time T and the larger N are, the better the accuracy. 2.2.5. We notice on Fig. 2.2 that the accuracy of the clock improves like N −1/2 , as N increases. For N = 106 and T = 0.9 s, the above formula gives 5.6 × 10−4 s. The hyperﬁne frequency of cesium is ω0 = 2π × 9.2 GHz, which corresponds to ∆ω/ω ∼ 10−14 . Section 2.3: The GPS System 2.3.1. One must see at least 4 satellites. With two of them, the diﬀerence between the two reception times t1 and t2 of the signals localize the observer on a surface (for instance on a plane at equal distances of the two satellites if t1 = t2 ); three satellites localize the observer on a line, and the fourth one determines the position of the observer unambiguously (provided of course that one assumes the observer is not deep inside the Earth or on a far lying orbit). 2.3.2. Suppose a satellite sends a signal at time t0 . This signal is received by an observer at a distance D at time t1 = t0 + D/c. If the clock of the satellite has drifted, the signal is not sent at time t0 , but at a slightly diﬀerent time t0 . The observer whom we assume has a correct time reference from another satellite, interprets the time t1 − t0 as a distance D = c(t1 − t0 ), he therefore makes an error c(t0 −t0 ) on his position. For a clock of relative accuracy 10−13 , the typical drift after 24 hours (=86 000 seconds) is 86 000 × 10−13 s, i.e. an error on the position of 2.5 meters. Note that the atomic clocks boarding the GPS satellites are noticeably less accurate than the fountain cold atom clocks in ground laboratories.

36

2 Atomic Clocks

Section 2.4: The Drift of Fundamental Constants Using the expression given in the text for the dependence on α of the frequencies ωCs and ωRb , we ﬁnd that a variation of the ratio R would be related to the variation of α by: 2 2 1 dα 11α2 ZCs − ZRb 1 dR = . R dt α dt 3 (1 + 11(αZRb )2 /6) (1 + 11(αZCs )2 /6) The quantity inside the brackets is 0.22, which leads to an upper bound of α/α ˙ of 1.4 × 10−14 per year, i.e. 4.3 × 10−22 per second. If we extrapolate this variation time to a time of the order of the age of the universe, this corresponds to a variation of 10−4 . Such an eﬀect should be detectable, in principle, by spectroscopic measurements on very far objects. Remark: a more precise determination of the α dependence of ωCs , for which the approximation Zα 1 is not very good, gives for the quantity inside the bracket a value of 0.45. Section 2.5: References The experimental data on the stability of a cold atom clock have been taken from the paper from the group of A. Clairon and C. Salomon, at Observatoire de Paris: G. Santarelli et al., Phys. Rev. Lett. 82, 4619 (1999). Concerning the drift of fundamental constants, see J. D. Prestage, R. L. Tjoelker, and L. Maleki, Phys. Rev. Lett. 74, 3511 (1995); H. Marion et al., Phys. Rev. Lett. 90, 150801 (2003); M. Fischer et al., Phys. Rev. Lett. 92, 230802 (2004).

3 Neutron Interferometry

In the late 1970s, Overhauser and his collaborators performed several neutron interference experiments which are of fundamental importance in quantum mechanics, and which settled debates which had started in the 1930s. We study in this chapter two of these experiments, aiming to measure the inﬂuence on the interference pattern (i) of the gravitational ﬁeld and (ii) of a 2π rotation of the neutron wave function. We consider here an interferometer made of three parallel, equally spaced crystalline silicon strips, as shown in Fig. 3.1. The incident neutron beam is assumed to be monochromatic.

Fig. 3.1. The neutron interferometer: The three “ears” are cut in a silicon monocrystal; C2 and C3 are neutron counters

For a particular value of the angle of incidence θ, called the Bragg angle, a plane wave ψinc = ei(p·r−Et)/¯h , where E is the energy of the neutrons and p their momentum, is split by the crystal into two outgoing waves which are symmetric with respect to the perpendicular direction to the crystal, as shown in Fig. 3.2.

38

3 Neutron Interferometry

Fig. 3.2. Splitting of an incident plane wave satisfying the Bragg condition

The transmitted wave and the reﬂected wave have complex amplitudes which can be written respectively as α = cos χ and β = i sin χ, where the angle χ is real: ψI = αei(p·r−Et)/¯h

ψII = βei(p ·r−Et)/¯h ,

(3.1)

where |p| = |p | since the neutrons scatter elastically on the nuclei of the crystal. The transmission and reﬂection coeﬃcients are T = |α|2 and R = |β|2 , with of course T + R = 1. In the interferometer shown in Fig. 3.1, the incident neutron beam is horizontal. It is split by the interferometer into a variety of beams, two of which recombine and interfere at point D. The detectors C2 and C3 count the outgoing neutron ﬂuxes. The neutron beam velocity corresponds to a de Broglie wavelength λ = 1.445 ˚ A. We recall the value of neutron mass M = 1.675×10−27 kg. The neutron beam actually corresponds to wave functions which are quasimonochromatic and which have a ﬁnite extension in the transverse directions. In order to simplify the writing of the equations, we only deal with pure monochromatic plane waves, as in (3.1).

3.1 Neutron Interferences 3.1.1. The measured neutron ﬂuxes are proportional to the intensities of the waves that reach the counters. Deﬁning the intensity of the incoming beam to be 1 (the units are arbitrary), write the amplitudes A2 and A3 of the wave functions which reach the counters C2 and C3 , in terms of α and β (it is not necessary to write the propagation terms ei(p·r−Et)/¯h ). Calculate the measured intensities I2 and I3 in terms of the coeﬃcients T and R. 3.1.2. Suppose that we create a phase shift δ of the wave propagating along AC, i.e. in C the wave function is multiplied by eiδ . (a) Calculate the new amplitudes A2 and A3 in terms of α, β and δ. (b) Show that the new measured intensities I2 and I3 are of the form I2 = µ − ν(1 + cos δ)

I3 = ν(1 + cos δ)

and express µ and ν in terms of T and R. (c) Comment on the result for the sum I2 + I3 .

3.2 The Gravitational Eﬀect

39

3.2 The Gravitational Eﬀect The phase diﬀerence δ between the beams ACD and ABD is created by rotating the interferometer by an angle φ around the direction of incidence. This creates a diﬀerence in the altitudes of BD and AC, which both remain horizontal, as shown in Fig. 3.3. The diﬀerence in the gravitational potential energies induces a gravitational phase diﬀerence. 3.2.1. Let d be the distance between the silicon strips, whose thickness is neglected here. Show that the side L of the lozenge ABCD and its height H, shown in Fig. 3.3, are related to d and to the Bragg angle θ by L = d/ cos θ and H = 2d sin θ. Experimentally the values of d and θ are d = 3.6 cm and θ = 22.1◦ .

Fig. 3.3. Turning the interferometer around the incident direction, in order to observe gravitational eﬀects

3.2.2. For an angle φ, we deﬁne the gravitational potential V to be V = 0 along AC and V = V0 along BD. (a) Calculate the diﬀerence ∆p of the neutron momenta in the beams AC and BD (use the approximation ∆p p). Express the result in terms of the momentum p along AC, the height H, sin φ, M , and the acceleration of gravity g. √ (b) Evaluate numerically the velocity 2gH. How good is the approximation ∆p p? 3.2.3. Evaluate the phase diﬀerence δ between the paths ABD and ACD. One can proceed in two steps: (a) Compare the path diﬀerence between the segments AB and CD. (b) Compare the path diﬀerence between the segments BD and AC. 3.2.4. The variation with φ of the experimentally measured intensity I2 in the counter C2 is represented in Fig. 3.4. (The data does not display a minimum exactly at φ = 0 because of calibration diﬃculties.) Deduce from these data the value of the acceleration due to gravity g.

40

3 Neutron Interferometry

Fig. 3.4. Measured neutron intensity in counter C2 as the angle φ is varied

3.3 Rotating a Spin 1/2 by 360 Degrees The plane of the setup is now horizontal. The phase diﬀerence arises by placing along AC a magnet of length l which produces a constant uniform magnetic ﬁeld B 0 directed along the z axis, as shown in Fig. 3.5.

Fig. 3.5. Experimental setup for observing the neutron spin Larmor precession

The neutrons are spin-1/2 particles, and have an intrinsic magnetic moˆ is the neutron spin operator, and the σ ˆ = µ0 σ ˆ where S ˆ = γn S ˆi ment µ (i = x, y, z) are the usual 2 × 2 Pauli matrices. The axes are represented in Fig. 3.5: the beam is along the y axis, the z axis is in the ABCD plane, and the x axis is perpendicular to this plane. We assume that the spin variables and the space variables are uncorrelated, i.e. at any point in space the wave function factorizes as

a+ (t) ψ+ (r, t) i(p·r−Et)/¯ h =e . ψ− (r, t) a− (t) We neglect any transient eﬀect due to the entrance and the exit of the ﬁeld zone.

3.3 Rotating a Spin 1/2 by 360 Degrees

41

The incident neutrons are prepared in the spin state

1 1 | + x = √ , 2 1 which is the eigenstate of µ ˆx with eigenvalue +µ0 . The spin state is not modiﬁed when the neutrons cross the crystal strips. 3.3.1. (a) Write the magnetic interaction Hamiltonian of the spin with the magnetic ﬁeld. (b) What is the time evolution of the spin state of a neutron in the magnet? h, calculate the three components of the expecta(c) Setting ω = −2µ0 B0 /¯ ˆ in this state, and describe the time evolution of µ ˆ in the tion value µ magnet. 3.3.2. When the neutron leaves the magnet, what is the probability Px (+µ0 ) of ﬁnding µx = +µ0 when measuring the x component of the neutron magnetic moment? For simplicity, one can set T = M lλ/(2π¯h) and express the result in terms of the angle δ = ωT /2. 3.3.3. For which values bn = nb1 (n integer) of the ﬁeld B0 is this probability equal to 1? To what motion of the average magnetic moment do these values bn correspond? A. Calculate b1 with µ0 = −9.65 × 10−27 J T−1 , l = 2.8 cm, λ = 1.445 ˚ 3.3.4. Write the state of the neutrons when they arrive on C2 and C3 (note p2 and p3 the respective momenta). 3.3.5. The counters C2 and C3 measure the neutron ﬂuxes I2 and I3 . They are not sensitive to spin variables. Express the diﬀerence of intensities I2 − I3 in terms of δ and of the coeﬃcients T and R. 3.3.6. The experimental measurement of I2 − I3 as a function of the applied ﬁeld B0 is given in Fig. 3.6. A numerical ﬁt of the curve shows that the distance between two maxima is ∆B = (64 ± 2) × 10−4 T.

Fig. 3.6. Diﬀerence of counting rates (I2 − I3 ) as a function of the applied ﬁeld

42

3 Neutron Interferometry

Comparing the values bn of question 3.3 with this experimental result, and recalling the result of a measurement of µx for these values, explain why this proves that the state vector of a spin-1/2 particle changes sign under a rotation by an odd multiple of 2π.

3.4 Solutions Section 3.1: Neutron Interferences 3.1.1. The beams ABDC2 and ACDC2 interfere. Omitting the propagation factors, one has, at C2 an amplitude A2 = α2 β + β 3 = β(α2 + β 2 ) . Similarly, for ABDC3 and ACDC3 , A3 = 2αβ 2 . The intensities at the two counters are I2 = R − 4R2 T

I3 = 4R2 T .

3.1.2. When there is a phase shift δ in C, the above expressions get modiﬁed as follows: A2 = α2 βeiδ + β 3 = β(α2 eiδ + β 2 ) ,

A3 = αβ 2 (1 + eiδ ) .

The intensities become I2 = R − 2R2 T (1 + cos δ)

I3 = 2R2 T (1 + cos δ) .

The fact that I2 + I3 does not depend on the phase shift δ is a consequence of the conservation of the total number of particles arriving at D. Section 3.2: The Gravitational Eﬀect 3.2.1. This results from elementary trigonometry. (a) Since there is no recoil energy of the silicon atoms to be taken care of, the neutron total energy (kinetic+potential) is a constant of the motion in all the 2 process. This energy is given by EAC = p2 /2M and EBD = (p − ∆p) /2M + M gH sin φ, hence ∆p M 2 gH sin φ/p . √ (b) The velocity 2gH is of the order of 0.5 m/s, and the neutron velocity is v = h/M λ 2700 m/s. The change in velocity ∆v is therefore very small: ∆v = gH/v 2 × 10−4 m/s for φ = π/2.

3.4 Solutions

43

3.2.2. (a) The gravitational potential varies in exactly the same way along AB and CD. The neutron state in both cases is a plane wave with momentum p = h/λ just before A or C. The same Schr¨odinger equation is used to determine the wave function at the end of the segments. This implies that the phases accumulated along the two segments AB and CD are equal. (b) When comparing the segments AC and BD, the previous reasoning does not apply, since the initial state of the neutron is not the same for the two segments. The initial state is exp(ipz/¯h) for AC, and exp[i(p − ∆p)z/¯h] for BD. After travelling over a distance L = AC = BD, the phase diﬀerence between the two paths is δ=

M 2 gλd2 ∆p L = tan θ sin φ . h ¯ π¯ h2

3.2.3. >From the previous result, one has δ2 − δ1 = A g (sin φ2 − sin φ1 ), h2 ). Therefore, where A = M 2 λd2 tan θ/(π¯ g=

δ2 − δ1 . A (sin φ2 − sin φ1 )

There are 9 oscillations, i.e. (δ2 − δ1 ) = 18π, between φ1 = −32◦ and φ2 = +24◦ , which gives g 9.8 ms−2 . The relative precision of the experiment was actually of the order of 10−3 . Section 3.3: Rotating a Spin 1/2 by 360 Degrees 3.3.1. Since B is along the z axis, the magnetic Hamiltonian is:

hω 1 0 ¯ ˆ HM = −µ · B 0 = . 0 −1 2 At time t, the spin state is 1 |Σ(t) = √ 2

e−iωt/2 e+iωt/2

.

d By a direct calculation of µ or by using Ehrenfest theorem ( dt µ = 1 ˆ ˆ [ µ, H]), we obtain: i¯ h

dµx = ωµy dt

dµy = −ωµx dt

dµz =0. dt

Initially µx = µ0 and µy = µz = 0; therefore, µ = µ0 (cos ωt ux + sin ωt uy ) .

44

3 Neutron Interferometry

3.3.2. When the neutrons leave the ﬁeld zone, the probability of ﬁnding µx = +µ0 is Px (+µ0 ) = |+x|Σ(T )|2 = cos2

ωT = cos2 δ 2

with T = l/v = lM λ/h. 3.3.3. The above probability is equal to 1 if δ = nπ (ωT = 2nπ), or B0 = nb1 with h2 2π 2 ¯ = 34.5 × 10−4 T . b1 = µ0 M lλ For δ = nπ the magnetic moment has rotated by 2nπ around the z axis by Larmor precession. 3.3.4. The formulas are similar to those found in question 1.2. The phase of the upper component of the spinor written in the {|+z , |−z } basis, is shifted by +δ, that of the lower component by −δ:

2 β β + α2 eiδ Amplitude at the counter C2 : ei(p2 ·r−Et)/¯h √ 2 2 −iδ 2 β +α e Amplitude at the counter C3 :

αβ 2 ei(p3 ·r−Et)/¯h √ 2

1 + eiδ 1 + e−iδ

3.3.5. Since the measuring apparatus is insensitive to spin variables, we must add the probabilities corresponding to Sz = ±1, each of which is the modulus squared of a sum of amplitudes. Altogether, we obtain the following intensities of the total neutron ﬂux in the two counters: I2 = R − 2R2 T (1 + cos δ) ,

I3 = 2R2 T (1 + cos δ)

and I2 − I3 = R − 4R2 T (1 + cos δ) . 3.3.6. There will be a minimum of I2 − I3 each time cos δ = +1, i.e. δ = 2nπ. This corresponds to a constructive interference in channel 3. On the other hand, there appears a maximum if cos δ = −1, i.e. δ = (2n + 1)π, and this corresponds to a destructive interference in channel 3 (I3 = 0). If δ = nπ, whatever the integer n, one is sure to ﬁnd the neutrons in the same spin state as in the initial beam. However, the interference pattern depends on the parity of n. The experimental result ∆B = (64 ± 2) × 10−4 T conﬁrms that if the spin has rotated by 4nπ, one recovers a constructive interference in channel 3 as in the absence of rotation, while if it has rotated by (4n + 2)π, the interference in C3 is destructive. The probability amplitude for the path ACD has changed sign in this latter case, although a spin measurement in this path after the magnet will give exactly the same result as on the incoming beam.

3.4 Solutions

45

References A.W. Overhauser, A.R. Collela, and S.A. Werner, Phys. Rev. Lett., 33, 1237 (1974); 34, 1472 (1975); 35, 1053 (1975). See also D. Greenberger and A.W. Overhauser, Scientiﬁc American, May 1980.

4 Spectroscopic Measurement on a Neutron Beam

We present here a very precise method for spectroscopic measurements, due to Norman Ramsey. The method, using atomic or molecular beams, can be applied to a very large class of problems. We shall analyse it in the speciﬁc case of a neutron beam, where it can be used to determine the neutron magnetic moment with high accuracy, by measuring the Larmor precession frequency in a magnetic ﬁeld B 0 . A beam of neutrons is prepared with velocity v along the x axis. The beam is placed in a constant uniform magnetic ﬁeld B 0 directed along the z axis. We write |+ and |− for the eigenstates of the z projection Sˆz of the neutron ˆ µ ˆ = γ S, ˆ being the spin, and γ for the gyromagnetic ratio of the neutron: µ ˆ neutron magnetic moment operator, and S its spin. The neutrons are initially in the state |−. When they approach the origin, they cross a zone where an oscillating ﬁeld B 1 (t) is applied in the (x, y) plane. The components of B1 are B1x = B1 e−r/a cos ω(t − z/c) B1y = B1 e−r/a sin ω(t − z/c) B1z = 0 ,

(4.1)

where r = x2 + y 2 . We assume that B1 is constant (strictly speaking it should vary in order to satisfy ∇ · B = 0) and that B1 B0 . In all parts of the chapter, the neutron motion in space is treated classically as a linear uniform motion. We are only interested in the quantum evolution of the spin state.

4.1 Ramsey Fringes 4.1.1. Consider a neutron whose motion in space is x = vt, y = 0, z = 0. ˆ What is the Hamiltonian H(t) describing the coupling of the neutron magnetic moment with the ﬁelds B 0 and B 1 ?

48

4 Spectroscopic Measurement on a Neutron Beam

Setting ω0 = −γB0 and ω1 = −γB1 , write the matrix representation of ˆ H(t) in the basis {|+, |−}. 4.1.2. Treating B 1 as a perturbation, calculate, in ﬁrst order time-dependent perturbation theory, the probability of ﬁnding the neutron in the state |+ at time t = +∞ (far from the interaction zone) if it was in the state |− at t = −∞. One measures the ﬂux of neutrons which have ﬂipped their spins, and are in the state |+ when they leave the ﬁeld zone. This ﬂux is proportional to the probability P−+ that they have undergone the above transition. Show that this probability has a resonant behavior as a function of the applied angular frequency ω. Plot P−+ as a function of the distance from the resonance ω − ω0 . How does the width of the resonance curve vary with v and a? The existence of this width puts a limit on the accuracy of the measurement of ω0 , and therefore of γ. Is there an explanation of this on general grounds? 4.1.3. On the path of the beam, one adds a second zone with an oscillating ﬁeld B 1 . This second zone is identical to the ﬁrst but is translated along the x axis by a distance b (b a):

= B1 e−r /a cos ω(t − z/c) B1x = B1 e−r /a sin ω(t − z/c) B1y =0, B1z

(4.2)

1/2

where r = ((x − b) + y 2 ) . Show that the transition probability P−+ across the two zones can be expressed in a simple way in terms of the transition probability calculated in the previous question. Why is it preferable to use a setup with two zones separated by a distance b rather than a single zone, as in question 4.1.2, if one desires a good accuracy in the measurement of the angular frequency ω0 ? What is the order of magnitude of the improvement in the accuracy? 2

4.1.4. What would be the probability P−+ if one were to use N zones equally spaced by the same distance b from one another? What optical system is this reminiscent of? 4.1.5. Suppose now that the neutrons, still in the initial spin state |−, propagate along the z axis instead of the x axis. Suppose that the length of the interaction zone is b, i.e. that the oscillating ﬁeld is given by (4.1) for −b/2 ≤ z ≤ +b/2 and is zero for |z| > b/2. Calculate the transition probability in this new conﬁguration. P−+ For what value of ω is this probability maximum? Explain the diﬀerence with the result obtained in question 4.1.2.

4.2 Solutions

49

4.1.6. In practice, the neutron beam has some velocity dispersion around the value v. Which of the two methods described in questions 4.1.3 and 4.1.5 is preferable? 4.1.7. Numerical Application: The neutrons of the beam have a de Broglie A. Calculate their velocity. wavelength λn = 31 ˚ In order to measure the neutron gyromagnetic ratio γn , one proceeds as in question 4.1.3. One can assume that the accuracy is given by δω0 =

πv . 2b

The most accurate value of the neutron gyromagnetic ratio is currently γn = −1.912 041 84 (±8.8 × 10−7 ) q/Mp where q is the unit charge and Mp the proton mass. In a ﬁeld B0 = 1 T, what must be the length b in order to achieve this accuracy?

4.2 Solutions 4.1.1. The magnetic Hamiltonian is ˆ ˆ H(t) = −µ.B = −γ B0 Sˆz + B1x (t) Sˆx + B1y (t) Sˆy . Since x = vt, y = z = 0, ˆ H(t) = −γ B0 Sˆz + B1 e−v|t|/a Sˆx cos ωt + Sˆy sin ωt whose matrix representation is

h ¯ ω1 exp(−v|t|/a − iωt) ω0 ˆ H(t) = . −ω0 2 ω1 exp(−v|t|/a + iωt) 4.1.2. Let |ψ(t) = α(t)|+ + β(t)|− be the neutron state at time t. The Schr¨ odinger equation gives the evolution of α and β: ω1 −iωt−v|t|/a ω0 α+ e β 2 2 ω0 ω1 iωt−v|t|/a e β. α− iβ˙ = 2 2

iα˙ =

˜ We now introduce the variables α ˜ and β: α ˜ (t) = α(t) eiω0 t/2 whose evolution is given by

˜ = β(t) e−iω0 t/2 , β(t)

50

4 Spectroscopic Measurement on a Neutron Beam

ω1 i(ω0 −ω)t−v|t|/a ˜ e β 2 ω1 i(ω−ω0 )t−v|t|/a ˙ iβ˜ = e α ˜. 2

iα ˜˙ =

The equation for α ˜ can be formally integrated and it gives ω1 t i(ω0 −ω)t −v|t |/a ˜ α ˜ (t) = β(t ) dt , e 2i −∞

(4.3)

where we have used the initial condition α ˜ (−∞) = α(−∞) = 0. Now, since ˜ ) by its we want the value of α(t) to ﬁrst order in B1 , we can replace β(t ˜ unperturbed value β(t ) = 1 in the integral. This gives ω1 +∞ i(ω0 −ω)t −v|t |/a e dt 2i −∞ 1 ω1 v = . ia (ω − ω0 )2 + (v/a)2

γ−+ ≡ α ˜ (+∞) =

The transition probability is therefore P−+ =

1 ω12 v 2 . a2 [(ω0 − ω)2 + (v/a)2 ]2

Fig. 4.1. Transition probability in one zone

The width of the resonance curve is of the order of v/a. This quantity is the inverse of the time τ = a/v a neutron spends in the oscillating ﬁeld. From the uncertainty relation δE.τ ∼ ¯ h, when an interaction lasts a ﬁnite time τ the accuracy of the energy measurement δE is bounded by δE ≥ ¯h/τ . Therefore, from ﬁrst principles, one expects that the resonance curve will have a width of the order of h ¯ /τ in energy, or 1/τ in angular frequency.

4.2 Solutions

51

4.1.3. In the two-zone case, the transition amplitude (in ﬁrst order perturbation theory) becomes

+∞ +∞ ω1 ei(ω0 −ω)t−v|t|/a dt + ei(ω0 −ω)t−|vt−b|/a dt . γ−+ = 2i −∞ −∞ If we make the change of variables t = t − b/v in the second integral, we obtain +∞ ω1 ei(ω0 −ω)t−v|t|/a dt , 1 + ei(ω0 −ω)b/v γ−+ = 2i −∞ which is the same formula as previously but multiplied by 1 + ei(ω0 −ω)b/v . If we square this expression, in order to ﬁnd the probability, we obtain

1 4ω12 v 2 (ω0 − ω)b 2 cos P−+ = . a2 [(ω0 − ω)2 + v 2 /a2 ]2 2v

Fig. 4.2. Ramsey fringes in a two-zone setup

The envelope of this curve is, up to a factor of 4, the same as the previous curve. However, owing to the extra oscillating factor, the half-width at halfmaximum of the central peak is now of order πv/(2b). The parameter which now governs the accuracy is the total time b/v that the neutron spends in the apparatus, going from one zone to the other. In spectroscopic measurements, it is important to locate the exact position of the maximum of the peak. Multiplying the width of the peak by a factor a/b ( 1 since a b) results in a major improvement of the measurement accuracy. Of course one could in principle build a single interaction zone of large size ∼ b, but it would be diﬃcult to maintain a well controlled oscillating ﬁeld over such a large region. From a practical point of view, it is much simpler to use small interaction zones of size a and to separate them by a large distance b.

52

4 Spectroscopic Measurement on a Neutron Beam

4.1.4. It is quite straightforward to generalize the previous results to an arbitrary number of zones: ω1 γ−+ = 1 + ei(ω0 −ω)b/v + · · · + ei(N −1)(ω0 −ω)b/v 2i +∞ ei(ω0 −ω)t−v|t|/a dt × −∞

P−+

1 sin2 [N (ω0 − ω)b/2v] ω2 v2 = 12 . 2 2 a [(ω0 − ω)2 + v 2 /a2 ] sin [(ω0 − ω)b/2v]

As far as amplitudes are concerned, there is a complete analogy with a diffraction grating in optics. The neutron (more generally, the particle or the atom) has some transition amplitude t for undergoing a spin ﬂip in a given interaction zone. The total amplitude T is the sum T = t + t eiφ + t e2iφ + . . . , where eiφ is the phase shift between two zones. 4.1.5. We now set z = vt, and x = y = 0 for the neutron trajectory. This will modify the phase of the ﬁeld (Doppler eﬀect) ω(t − z/c) → ω(1 − v/c)t = ω ˜t

with

ω ˜ = ω (1 − v/c)

and we must integrate the evolution of α: ˜ iα ˜˙ =

ω1 i(ω0 −˜ω)t ˜ e β 2

(with β˜ 1) between ti = −b/(2v) and tf = b/(2v). The transition probability is then sin2 [(ω0 − ω ˜ )b/(2v)] = ω12 , P−+ 2 (ω0 − ω ˜) which has a width of the order of b/v but is centered at ω ˜ = ω0 ⇒ ω =

ω0 ω0 (1 + v/c) . 1 − v/c

Comparing with question 4.1.2, we ﬁnd that the resonance frequency is displaced: The neutron moves in the propagation direction of the ﬁeld, and there is a ﬁrst order Doppler shift of the resonance frequency. 4.1.6. If the neutron beam has some velocity dispersion, the experimental result will be the same as calculated above, but smeared over the velocity distribution. In the method of question 4.1.3, the position of lateral fringes, and the width of the central peak, vary with v. A velocity distribution will lead to

4.2 Solutions

53

a broader central peak and lateral fringes of decreasing amplitude. However the position of the central peak does not depend on the velocity, and it is therefore not shifted if the neutron beam has some velocity dispersion. On the contrary, in the method of question 4.1.5, the position of the central peak depends directly on the velocity. A dispersion in v will lead to a corresponding dispersion of the position of the peak we want to measure. The ﬁrst method is highly preferable. A, v = h/(Mn λn ) 128 m/s. 4.1.7. Numerically, for λn = 31 ˚ Experimentally, one obtains an accuracy δω0 /ω0 = δγn /γn = 4.6 × 10−7 . For B = 1 T, the angular frequency is ω0 = γn B0 1.8 × 108 s−1 , which gives δω0 /(2π) 13 Hz and b 2.4 m. Actually, one can improve the accuracy considerably by analysing the shape of the peak. In the experiment reported in the reference quoted below, the length b is 2 m and the ﬁeld is B0 = 0.05 T (i.e. an angular frequency 20 times smaller than above). Reference G.L. Greene, N.F. Ramsey, W. Mampe, J.M. Pendlebury, K.F. Smith, W.D. Dress, P.D. Miller, and P. Perrin, Phys. Rev. D 20, 2139 (1979).

5 Analysis of a Stern–Gerlach Experiment

We analyze a Stern–Gerlach experiment, both experimentally and from the theoretical point of view. In the experimental setup considered here, a monochromatic beam of neutrons crosses a region of strongly inhomogeneous magnetic ﬁeld, and one observes the outgoing beam.

5.1 Preparation of the Neutron Beam Neutrons produced in a reactor are ﬁrst “cooled”, i.e. slowed down by crossing liquid hydrogen at 20 K. They are incident on a monocrystal, for instance graphite, from which they are diﬀracted. To each outgoing direction, there corresponds a well-deﬁned wavelength, and therefore a well-deﬁned momentum. A beryllium crystal acts as a ﬁlter to eliminate harmonics, and the vertical extension of the beam is controlled by two gadolinium blocks, which are opaque to neutrons, separated by a thin sheet of (transparent) aluminum of thickness a, which constitutes the collimating slit, as shown in Fig. 5.1.

Fig. 5.1. Preparation of the neutron beam

5.1.1. The de Broglie wavelength of these monochromatic neutrons is λ = 4.32 ˚ A. What are their velocity and their kinetic energy?

56

5 Analysis of a Stern–Gerlach Experiment

5.1.2. One observes the impacts of the neutrons on a detector at a distance L = 1 m from the slit. The vertical extension of the beam at the detector is determined by two factors, ﬁrst the width a of the slit, and second the diﬀraction of the neutron beam by the slit. We recall that the angular width θ of the diﬀraction peak from a slit of width a is related to the wavelength λ by sin θ = λ/a. For simplicity, we assume that the neutron beam is well collimated before the slit, and that the vertical extension δ of the beam on the detector is the sum of the width a of the slit and the width of the diﬀraction peak. Show that one can choose a in an optimal way in order to make δ as small as possible. What is the corresponding width of the beam on the detector? 5.1.3. In the actual experiment, the chosen value is a = 5 µm. What is the observed width of the beam at the detector? Comment on the respective eﬀects of the slit width a and of diﬀraction, on the vertical shape of the observed beam on the detector? The extension of the beam corresponds to the distribution of neutron impacts along the z axis. Since the purpose of the experiment is not only to observe the beam, but also to measure its “position” as deﬁned by the maximum of the distribution, what justiﬁcation can you ﬁnd for choosing a = 5 µm? Figure 5.2 is an example of the neutron counting rate as a function of z. The horizontal error bars, or bins, come from the resolution of the measuring apparatus, the vertical error bars from the statistical ﬂuctuations of the number of neutrons in each bin. The curve is a best ﬁt to the experimental points. Its maximum is determined with an accuracy δz ∼ 5 µm.

Fig. 5.2. Measurement of the beam proﬁle on the detector

5.3 The Stern–Gerlach Experiment

57

5.2 Spin State of the Neutrons In order to completely describe the state of a neutron, i.e. both its spin state and its spatial state, we consider the eigenbasis of the spin projection along the z axis, Sˆz , and we represent the neutron state as

ψ+ (r, t) , |ψ(t) : ψ− (r, t) where the respective probabilities of ﬁnding the neutron in the vicinity of h/2 are point r with its spin component Sz = ±¯ d3 P (r, Sz = ±¯ h/2, t) = |ψ± (r, t)|2 d3 r . 5.2.1. What are the probabilities P± (t) of ﬁnding, at time t, the values ±¯ h/2 when measuring Sz irrespective of the position r? 5.2.2. What is, in terms of ψ+ and ψ− , the expectation value of the x component of the neutron spin Sx in the state |ψ(t)? 5.2.3. What are the expectation values of the neutron’s position r and momentum p in the state |ψ(t)? 5.2.4. We assume that the state of the neutron can be written:

α+ |ψ(t) : ψ(r, t) , α− where the two complex numbers α± are such that |α+ |2 + |α− |2 = 1. How do the results of questions 5.2.2 and 5.2.3 simplify in that case?

5.3 The Stern–Gerlach Experiment Between the slit, whose center is located at the origin (x = y = z = 0), and the detector, located in the plane x = L, we place a magnet of length L whose ﬁeld B is directed along the z axis. The magnetic ﬁeld varies strongly with z; see Fig. 5.3. We assume that the components of the magnetic ﬁeld are Bx = By = 0

Bz = B0 + b z .

In what follows we choose1 B0 = 1 T and b = 100 T/m. 1

This form violates Maxwell’s equation ∇ · B = 0, but it simpliﬁes the following calculation. With a little modiﬁcation (e.g. Bx = 0, By = −b y and By Bz over the region of space crossed by the neutron beam), one can settle this matter, and arrive at the same conclusions.

58

5 Analysis of a Stern–Gerlach Experiment

Fig. 5.3. Magnetic ﬁeld setup in the Stern–Gerlach experiment

ˆ in the matrix representation that The magnetic moment of the neutron µ we have chosen for |ψ is ˆ = µ0 σ ˆ , µ ˆ are the usual Pauli matrices, and µ0 = 1.913 µN , where µN is the where σ h/2Mp = 5.051 10−27 J · T−1 . Hereafter, we denote nuclear magneton µN = q¯ the neutron mass by m. 5.3.1. What is the form of the Hamiltonian for a neutron moving in this magnetic ﬁeld? Write the time-dependent Schr¨ odinger equation for the state |ψ(t). Show that the Schr¨ odinger equation decouples into two equations of the Schr¨ odinger type, for ψ+ and ψ− respectively. 5.3.2. Show that one has d dt

|ψ± (r, t)|2 d3 r = 0 .

What does one conclude as to the probabilities of measuring µz = ±µ0 ? 5.3.3. We assume that, at t = 0, at the entrance of the ﬁeld zone, one has

α+ |ψ(0) : ψ(r, 0) α− and that r = 0, py = pz = 0 and px = p0 = h/λ, where the value of the wavelength λ has been given above. The above conditions correspond to the experimental preparation of the neutron beam discussed in Sect. 5.1. Let Aˆ be an observable depending on the position operator rˆ and the ˆ . We deﬁne the numbers A+ and A− by momentum operator p 1 ∗ ∗ A± = (r, t) Aˆ ψ± (r, t) d3 r . ψ± |α± |2

5.4 Solutions

59

What is the physical interpretation of A+ and A− ? Show in particular that |ψ+ |2 /|α+ |2 and |ψ− |2 /|α− |2 are probability laws. 5.3.4. Apply Ehrenfest’s theorem to calculate the following quantities: d r ± , dt

d p± . dt

Solve the resulting equations and give the time evolution of r ± and p± . Give the physical interpretation of the result, and explain why one observes a splitting of the initial beam into two beams of relative intensities |α+ |2 and |α− |2 . 5.3.5. Calculate the splitting between the two beams when they leave the magnet. Express the result in terms of the kinetic energy of the incident neutrons (we recall that L = 1 m and b = 100 T/m). Given the experimental error δz in the measurement of the position of the maximum intensity of a beam, i.e. δz = 5 × 10−6 m as discussed in question 5.1.3, what is the accuracy on the measurement of the neutron magnetic moment in such an experiment, assuming that the determination of the magnetic ﬁeld and the neutron energy is not a limitation? Compare with the result of magnetic resonance experiments: µ0 = (−1.91304184 ± 8.8 × 10−7 ) µN . 5.3.6. In the same experimental setup, what would be the splitting of a beam of silver atoms (in the original experiment of Stern and Gerlach, the atomic beam came from an oven at 1000 K) of energy E 1.38 × 10−20 J? The magnetic moment of a silver atom is the same as that of the valence electron h/2me 9.3 × 10−24 J.T−1 . |µe | = q¯ 5.3.7. Show that, quite generally, in order to be able to separate the two outgoing beams, the condition to be satisﬁed is of the form E⊥ t ≥ ¯ h/2 , where E⊥ is the transverse kinetic energy acquired by the neutrons in the process, and t is the time they spend in the magnetic ﬁeld. Comment and conclude.

5.4 Solutions Section 5.1: Preparation of the Neutron Beam 5.1.1. We have v = h/(λm) and E = mv 2 /2, which yields v = 916 m s−1 and E = 0.438 × 10−2 eV.

60

5 Analysis of a Stern–Gerlach Experiment

5.1.2. The contribution of diﬀraction to the beam width is δdiﬀ = L tan θ ∼ Lλ/a. With the simple additive prescription (which can be improved, but this would not yield √ very diﬀerent results), we obtain δ = a + Lλ/a which is minimal for a = Lλ 21 µm. The spreading√of the beam on the detector is then equal to the Heisenberg minimum δ = 2 Lλ = 42 µm. The uncertainty relations forbid δ to be less than some lower limit. In other words, the spreading of the wave packet, which increases as a decreases competes with the spatial deﬁnition of the incoming beam. 5.1.3. For a = 5 µm, we have δ = 91.5 µm. In that case, the eﬀect of diﬀraction is predominant. The reason for making this choice is that the shape of the diﬀraction peak is known and can be ﬁtted quite nicely. Therefore, this is an advantage in determining the position of the maximum. However, one cannot choose a to be too small, otherwise the neutron ﬂux becomes too small, and the number of events is insuﬃcient. Section 5.2: Spin State of the Neutrons 5.2.1. P± (t) = |ψ± (r, t)|2 d3 r N.B. The normalization condition (total probability equal to 1) is

P+ + P − = 1 ⇒ |ψ+ (r, t)|2 + |ψ− (r, t)|2 d3 r = 1 . The quantity |ψ+ (r, t)|2 + |ψ− (r, t)|2 is the probability density of ﬁnding the neutron at point r. h/2)ψ|ˆ σx |ψ 5.2.2. By deﬁnition, the expectation value of Sx is Sx = (¯ therefore ∗

h ¯ ∗ ψ+ (r, t)ψ− (r, t) + ψ− Sx = (r, t)ψ+ (r, t) d3 r . 2 5.2.3. Similarly

r = r |ψ+ (r, t)|2 + |ψ− (r, t)|2 d3 r ,

∗ h ¯ ∗ p = (r, t) ∇ψ− (r, t) d3 r . ψ+ (r, t) ∇ψ+ (r, t) + ψ− i 5.2.4. If the variables are factorized, we have the simple results:

∗ Sx = h ¯ Re α+ α−

and r =

r |ψ(r, t)|2 d3 r ,

p =

¯ h i

ψ ∗ (r, t) ∇ψ(r, t) d3 r .

5.4 Solutions

61

Section 5.3: The Stern–Gerlach Experiment 5.3.1. The matrix form of the Hamiltonian is

ˆ2 1 0 p 1 0 ˆ H= . − µ0 (B0 + b zˆ) 0 −1 2m 0 1 The Schr¨ odinger equation is i¯ h

d ˆ |ψ(t) = H|ψ(t) . dt

If we write it in terms of the coordinates ψ± we obtain the uncoupled set h2 ¯ ∂ ψ+ (r, t) = − ∆ψ+ − µ0 (B0 + b z) ψ+ ∂t 2m ∂ h2 ¯ ∆ψ− + µ0 (B0 + b z) ψ− i¯ h ψ− (r, t) = − ∂t 2m i¯ h

d ˆ ± |ψ± , with or, equivalently i¯ h dt |ψ± = H 2

h ˆ± = − ¯ H ∆ ∓ µ0 (B0 − b z) . 2m In other words, we are dealing with two uncoupled Schr¨ odinger equations, where the potentials have opposite values. This is basically what causes the Stern–Gerlach splitting. ˆ± 5.3.2. Since both ψ+ and ψ− satisfy Schr¨ odinger equations, and since H are both hermitian, we have the usual properties of Hamiltonian evolution for ψ+ and ψ− separately, in particular the conservation of the norm. The probability of ﬁnding µz = ±µ0 , and the expectation value of µz are both time independent. 5.3.3. By deﬁnition, we have |ψ± (r, t)|2 d3 r = |α± |2 , where |α± |2 is time independent. The quantities |ψ+ (r, t)|2 /|α+ |2 and |ψ− (r, t)|2 /|α− |2 are the probability densities for ﬁnding a neutron at poh/2 and Sz = −¯ h/2. sition r with, respectively, Sz = +¯ The quantities A+ and A− are the expectation values of the physical h/2 and Sz = −¯ h/2. quantity A, for neutrons which have, respectively, Sz = +¯ 5.3.4. Applying Ehrenfest’s theorem, one has for any observable d 1 ∗ ˆ H ˆ ± ] ψ± (r, t) d3 r . A± = (r, t) [A, ψ± dt i¯ h|α± |2

62

5 Analysis of a Stern–Gerlach Experiment

Therefore

d r ± = p± /m dt

and

d d d px± = py± = 0 pz± = ±µ0 b . dt dt dt The solution of these equations is straightforward: px± = p0 ,

py± = 0 ,

pz± = ±µ0 b t

p0 t µ0 b t2 = vt , y± = 0 , z± = ± . m 2m Consequently, the expectation values of the vertical positions of the neutrons which have µz = +µ0 and µz = −µ0 diverge as time progresses: there is a separation in space of the support of the two wave functions ψ+ and ψ− . The intensities of the two outgoing beams are proportional to |α+ |2 and |α− |2 . x± =

5.3.5. As the neutrons leave the magnet, one has x = L, therefore t = L/v and ∆z = |µ0 b |L2 /mv 2 = |µ0 b |L2 /2E where E is the energy of the incident neutrons. This provides a splitting of ∆z = 0.69 mm. The error on the position of each beam is δz = 5 µm, that is to say a relative error on the splitting of the beams, or equivalently, on the measurement of µ0 √ δµ0 2 δz ∼ 1.5% , µ0 ∆z which is far from the accuracy of magnetic resonance measurements. 5.3.6. For silver atoms, one has |µ0 |/2E = 3.4 × 10−4 T−1 . Hence, in the same conﬁguration, one would obtain, for the same value of the ﬁeld gradient and the same length L = 1 m, a separation ∆z = 3.4 cm, much larger than for neutrons. Actually, Stern and Gerlach, in their ﬁrst experiment, had a much weaker ﬁeld gradient and their magnet was 20 cm long. 5.3.7. The condition to be satisﬁed in order to resolve the two outgoing beams is that the distance ∆z between the peaks should be larger than the full width of each peak (this is a common criterion in optics; by an appropriate inspection of the line shape, one may lower this limit). We have seen in Sect. 5.1 that √ the absolute minimum for the total beam extension√on the detector is 2 Lλ, which amounts to a full width at half-maximum Lλ. In other words, we must have: ∆z 2 ≥ L λ . In the previous section, we have obtained the value of ∆z, and, by squaring, we 2 obtain ∆z 2 = (µ0 b ) t4 /m2 , where t is the time spent traversing the magnet. On the other hand, the transverse kinetic energy of an outgoing neutron is 2 E⊥ = p2z± /(2m) = (µ0 b ) t2 /(2m).

5.4 Solutions

63

Putting the two previous relations together, we obtain ∆z 2 = 2E⊥ t2 /m; inserting this result in the ﬁrst inequality, we obtain E⊥ t ≥ h/2 , where we have used L = vt and λ = h/mv. This is nothing but one of the many forms of the time–energy uncertainty relation. The right-hand side is not the standard h ¯ /2 because we have considered a rectangular shape of the incident beam (and not a Gaussian). This brings in an extra factor of 2π. Physically, this result is interesting in many respects. (a) First, it shows that the eﬀort that counts in making the experiment feasible is not to improve individually the magnitude of the ﬁeld gradient, or the length of the apparatus, etc., but the particular combination of the product of the energy transferred to the system and the interaction time of the system with the measuring apparatus. (b) Secondly, this is a particular example of the fundamental fact stressed by many authors2 that a measurement is never point-like. It has always a ﬁnite extension both in space and in time. The Stern–Gerlach experiment is actually a very good example of a measuring apparatus in quantum mechanics since it transfers quantum information – here the spin state of the neutron – into space–time accessible quantities – here the splitting of the outgoing beams. (c) This time–energy uncertainty relation is encountered in most, if not all quantum measurements. Here it emerges as a consequence of the spreading of the wave packet. It is a simple and fruitful exercise to demonstrate rigorously the above property by calculating directly the time evolution of the following expectation values: 2 pz± 2 − z± 2 , ET = , z± p± + p± z± . z± , ∆z 2 = z± 2m

2

See, for instance, L.D. Landau and E.M. Lifshitz, Quantum Mechanics, Pergamon Press, Oxford, 1965.

6 Measuring the Electron Magnetic Moment Anomaly

In the framework of the Dirac equation, the gyromagnetic factor g of the electron is equal to 2. In other words, the ratio between the magnetic moment and the spin of the electron is gq/(2m) = q/m, where q and m are the charge and the mass of the particle. When one takes into account the interaction of the electron with the quantized electromagnetic ﬁeld, one predicts a value of g slightly diﬀerent from 2. The purpose of this chapter is to study the measurement of the quantity g − 2.

6.1 Spin and Momentum Precession of an Electron in a Magnetic Field Consider an electron, of mass m and charge q (q < 0), placed in a uniform and static magnetic ﬁeld B directed along the z axis. The Hamiltonian of the electron is ˆ 2−µ ˆ = 1 (ˆ ˆ · B, p − q A) H 2m ˆ is the vector potential A ˆ = B × rˆ /2 and µ ˆ is the intrinsic magnetic where A moment operator of the electron. This magnetic moment is related to the ˆ by µ ˆ with γ = (1 + a)q/m. The quantity a is called ˆ = γ S, spin operator S the magnetic moment “anomaly”. In the framework of the Dirac equation, a = 0. Using quantum electrodynamics, one predicts at ﬁrst order in the ﬁne structure constant a = α/(2π). ˆ ˆ = (ˆ The velocity operator is v p − q A)/m, and we set ω = qB/m. 6.1.1. Verify the following commutation relations: ˆ = i¯ hω vˆy ; [ˆ vx , H]

ˆ = −i¯ [ˆ vy , H] hω vˆx ;

ˆ =0. [ˆ vz , H]

6.1.2. Consider the three quantities C1 (t) = Sˆz vˆz ,

C2 (t) = Sˆx vˆx + Sˆy vˆy ,

C3 (t) = Sˆx vˆy − Sˆy vˆx .

66

6 Measuring the Electron Magnetic Moment Anomaly

Write the time evolution equations for C1 , C2 , C3 . Show that these three equations form a linear diﬀerential system with constant coeﬃcients. One will make use of the quantity Ω = a ω. ˆ ·v ˆ ? 6.1.3. What is the general form for the evolution of S 6.1.4. A beam of electrons of velocity v is prepared at time t = 0 in a spin state such that one knows the values of C1 (0), C2 (0), and C3 (0). The beam interacts with the magnetic ﬁeld B during the time interval [0, T ]. One neglects the interactions between the electrons of the beam. At time T , one ˆ v . measures a quantity which is proportional to S.ˆ The result of such a measurement is presented in Fig. 6.1 as a function of the time T , for a value of the magnetic ﬁeld B = 9.4 × 10−3 T (data taken from D.T. Wilkinson and H.R. Crane, Phys. Rev. 130, 852 (1963)). Deduce from this curve an approximate value for the anomaly a. 6.1.5. Does the experimental value agree with the prediction of quantum electrodynamics?

ˆ v , as a function of the time T Fig. 6.1. Variations of the quantity S.ˆ

6.2 Solutions ˆ = mˆ 6.1.1. The electron Hamiltonian is H v 2 /2 − γB Sˆz . The following commutation relations can be established with no diﬃculty hqB/m2 = i¯ hω/m, [ˆ vx , vˆz ] = [ˆ vy , vˆz ] = 0 , [ˆ vx , vˆy ] = i¯ 2 [ˆ vx , vˆy ] = [ˆ vx , vˆy ]ˆ vy + vˆy [ˆ vx , vˆy ] = 2i¯ hω vˆy /m . Therefore ˆ = i¯ [ˆ vx , H] hωˆ vy ;

ˆ = −i¯ [ˆ vy , H] hωˆ vx ;

ˆ =0. [ˆ vz , H]

6.2 Solutions

67

ˆ = [O, ˆ H], ˆ 6.1.2. We make use of the property i¯h(d/dt)O valid for any observable (Ehrenfest theorem). The time evolution of C1 is trivial: ˆ =0 ⇒ [Sˆz vˆz , H]

dC1 =0; dt

C1 (t) = A1 ,

where A1 is a constant. For C2 and C3 , we proceed in the following way: ˆ = [Sˆx vˆx , mˆ [Sˆx vˆx , H] v 2 /2] − γB[Sˆx vˆx , Sˆz ] = i¯ hω(Sˆx vˆy + (1 + a)Sˆy vˆx ). Similarly, ˆ = −i¯ [Sˆy vˆy , H] hω(Sˆy vˆx + (1 + a)Sˆx vˆy ) ˆ = −i¯ [Sˆx vˆy , H] hω(Sˆx vˆx − (1 + a)Sˆy vˆy ) ˆ = i¯ [Sˆy vˆx , H] hω(Sˆy vˆy − (1 + a)Sˆx vˆx ) . Therefore, ˆ = −i¯ [Sˆx vˆx + Sˆy vˆy , H] hωa (Sˆx vˆy − Sˆy vˆx ) ˆ = i¯ [Sˆx vˆy − Sˆy vˆx , H] hωa(Sˆx vˆx + Sˆy vˆy ) and

dC2 = −ΩC3 , dt

dC3 = ΩC2 . dt

6.1.3. We therefore obtain d2 C2 /dt2 = −Ω 2 C2 , whose solution is C2 (t) = A2 cos (Ωt + ϕ) , where A2 and ϕ are constant. Hence, the general form of the evolution of S · v is S · v(t) = C1 (t) + C2 (t) = A1 + A2 cos (Ωt + ϕ) . In other words, in the absence of anomaly, the spin and the momentum of the electron would precess with the same angular velocity: the cyclotron frequency (precession of momentum) and the Larmor frequency (precession of magnetic moment) would be equal. Measuring the diﬀerence in these two frequencies gives a direct measurement of the anomaly a, of fundamental importance in quantum electrodynamics. 6.1.4. One calculates the anomaly from the relation a = Ω/ω. The experimental results for S · v show a periodic behavior in time with a period τ ∼ 3 µs, i.e. Ω = 2π/τ ∼ 2 × 106 s−1 . In a ﬁeld B = 0.0094 T, ω = 1.65 × 109 s−1 , and a = Ω/ω ∼ 1.2 × 10−3 . 6.1.5. This value is in good agreement with the theoretical prediction a = α/2π = 1.16 × 10−3 .

68

6 Measuring the Electron Magnetic Moment Anomaly

Remark: The value of the anomaly is now known with an impressive accuracy: atheo. = 0.001 159 652 200 (40) aexp. = 0.001 159 652 193 (10) . The theoretical calculation includes all corrections up to order 3 in α.

7 Decay of a Tritium Atom

The nucleus of the tritium atom is the isotope 3 H, of charge Z = 1. This nucleus is radioactive and transforms into a 3 He nucleus by β decay. The purpose of this chapter is to study the electronic state of the 3 He+ ion formed after the decay. We consider nuclei as inﬁnitely massive compared to the electron, of ¯ 2 /(me2 ) for the Bohr radius and EI = mc2 α2 /2 mass m. We write a1 = h 13.6 eV for the ionization energy of the hydrogen atom, where α is the ﬁne structure constant [e2 = q 2 /(4π0 ), where q is the electron charge]. In the ground state |ψ0 of the tritium atom, the wave function of the electron (n = 1, l = 0, m = 0) is the same as in the normal hydrogen atom: 1 ψ0 (r) = √ e−r/a1 . (7.1) πa1 3 The β decay of the tritium nucleus leads to: 3

H → 3 He + e− + ν¯

(7.2)

(¯ ν is an antineutrino), where the emitted electron has an energy of the order of 15 keV and the helium nucleus 3 He has charge Z = 2. The decay is an instantaneous process; the β electron is emitted with a large velocity and leaves the atomic system very rapidly. Consequently, an ionized 3 He+ atom is formed, for which, at the time t0 of the decay, the wave function of the electron is practically the same as in tritium, and we shall assume it is still given by (7.1). We denote by |n, l, m the states of the ionized helium atom which is a hydrogen-like system, i.e. one electron placed in the Coulomb ﬁeld of a nucleus of charge 2.

7.1 The Energy Balance in Tritium Decay ˆ 1 of the atomic electron before the decay and 7.1.1. Write the Hamiltonian H ˆ the Hamiltonian H2 of this electron after the decay (when the potential term has suddenly changed).

70

7 Decay of a Tritium Atom

7.1.2. What are, in terms of EI , the energy levels of the 3 He+ atom? Give its Bohr radius and its ground state wave function ϕ100 (r). 7.1.3. Calculate the expectation value E of the energy of the electron after the decay. One can for instance make use of the fact that: ψ0 |

1 1 |ψ0 = r a1

2 ˆ2 = H ˆ1 − e . H r

and

Give the value of E in eV. 7.1.4. Express in terms of |ψ0 and |n, l, m the probability amplitude c(n, l, m) and the probability p(n, l, m) of ﬁnding the electron in the state |n, l, m of 3 He+ after the decay. Show that only the probabilities pn = p (n, 0, 0) do not vanish. 7.1.5. Calculate the probability p1 of ﬁnding the electron in the ground state of 3 He+ . What is the corresponding contribution to E? 7.1.6. A numerical calculation gives the following values: p2 =

1 , 4

∞

pn = 0.02137,

n=3

∞ pn = 0.00177. 2 n n=3

∞ Calculate the probability n=1 pn of ﬁnding the atomic electron in a bound state of 3 He+ and the corresponding contribution to E. Comment on the result. 7.1.7. Experimentally, in the β decay of the tritium atom, one observes that, in about 3% of the events, there are two outgoing electrons, one with a mean kinetic energy Ek ∼ 15 keV, the other with Ek ∼ 34.3 eV, thus leaving a completely ionized 3 He2+ nucleus, as if the β decay electron had “ejected” the atomic electron. Explain this phenomenon.

7.2 Solutions 7.1.1. The two Hamiltonians are 2 2 ˆ 1 = pˆ − e H 2m r

2 2 ˆ 2 = pˆ − 2e . H 2m r

7.1.2. The energy levels corresponding to the bound states of a hydrogenlike atom of nuclear charge Z are En = −Z 2 EI /n2 . In the present case, En = −4EI /n2 . The new Bohr radius is a2 = a1 /2, and the wave function is 1 ϕ100 (r) = 3 e−r/a2 . πa2

7.3 Comments

71

7.1.3. The expectation value of the electron energy in the new nuclear conﬁguration is 2 ˆ 2 |ψ0 = ψ0 |H ˆ 1 |ψ0 − ψ0 | e |ψ0 , E = ψ0 |H r which amounts to

E = −EI −

e2 = −3EI −40.8 eV . a1

7.1.4. By deﬁnition, the probability amplitude is c(n, l, m) = n, l, m|ψ0 , and the probability p(n, l, m) = |n, l, m|ψ0 |2 . The analytic form is ∗ c(n, l, m) = Rnl (r) (Yl,m (θ, φ)) ψ0 (r) d3 r , where Rnl (r) are the radial wave functions of the 3 He+ hydrogen-like atom. Since ψ0 is of the form ψ0 (r) = χ(r)Y0,0 (θ, φ), the orthogonality of spherical harmonics implies p(n, l, m) = 0 if (l, m) = (0, 0). 7.1.5. The probability amplitude in the lowest energy state is √ −r/a2 −r/a1 e e 16 2 1/2 2 (p1 ) . = 4π r dr = 27 πa32 πa31 Hence the probability p1 = 0.70233 and the contribution to the energy p1 E1 = −38.2 eV. 7.1.6. With the numerical ∞ values given in the text, one has p2 E2 = −EI /4 = −3.4 eV, and p = 1 pn = 0.9737 . The contribution to E is EB = ∞ p E = −3.0664 E n n I = −41.7 eV. 1 The total probability is smaller than 1; there exists a non-zero probability (1 − p) = 0.026 that the atomic electron is not bound in the ﬁnal state. The contribution of bound states EB = −41.7 eV is smaller than the total expectation value of the energy E by 0.9 eV. The probability (1 − p) corresponds therefore to a positive electron energy, i.e. an ionization of 3 He+ into 3 He2+ with emission of the atomic electron. 7.1.7. There is necessarily a probability 1 − p = 0.026 for the atomic electron not to be bound around the helium nucleus, therefore that the helium atom be completely ionized in the decay. If the mean kinetic energy of the expelled electron is Ek ∼ 34.3 eV, this represents a contribution of the order of (1 − p)Ek ∼ +0.89 eV to the mean energy which compensates the apparent energy deﬁcit noted above.

7.3 Comments This type of reaction is currently being studied in order to determine the neutrino mass. If M1 and M2 are the masses of the two nuclei, Eβ the energy

72

7 Decay of a Tritium Atom

of the β electron, E the energy of the atomic electron, and Eν¯ the neutrino energy, energy conservation gives for each event: M1 c2 − EI = M2 c2 + Eβ + Eν¯ + E. For a given value of E, the determination of the maximum energy of the β electron (which covers all the spectrum up to 19 keV in the tritium atom case) provides a method for determining the minimum value mν¯ c2 of Eν¯ through this energy balance. An important theoretical problem is that current experiments are performed on molecular tritium (HT or TT molecules) and that molecular wave functions are not known explicitly, contrary to the atomic case considered here. The most precise experiment up to date is reported in Weinheimer et al., Phys. Lett. B460, 219, (1999).

8 The Spectrum of Positronium

The positron e+ is the antiparticle of the electron. It is a spin-1/2 particle, which has the same mass m as the electron, but an electric charge of opposite sign. In this chapter we consider the system called positronium which is an atom consisting of an e+ e− pair.

8.1 Positronium Orbital States We ﬁrst consider only the spatial properties of the system, neglecting all spin eﬀects. We only retain the Coulomb interaction between the two particles. No proof is required, an appropriate transcription of the hydrogen atom results suﬃces. 8.1.1. Express the reduced mass of the system µ, in terms of the electron mass m. 8.1.2. Write the Hamiltonian of the relative motion of the two particles in terms of their separation r and their relative momentum p. 8.1.3. What are the energy levels of the system, and their degeneracies? How do they compare with those of hydrogen? 8.1.4. What is the Bohr radius a0 of the system? How do the sizes of hydrogen and positronium compare? 8.1.5. Give the expression for the normalized ground state wave function ¯, ψ100 (r). Express |ψ100 (0)|2 in terms of the fundamental constants: m, c, h and the ﬁne structure constant α.

8.2 Hyperﬁne Splitting We now study the hyperﬁne splitting of the ground state.

74

8 The Spectrum of Positronium

8.2.1. What is the degeneracy of the orbital ground state if one takes into account spin variables (in the absence of a spin–spin interaction)? 8.2.2. Explain why the (spin) gyromagnetic ratios of the positron and of the electron have opposite signs: γ1 = −γ2 = γ. Express γ in terms of q and m. 8.2.3. One assumes that, as in hydrogen, the spin–spin Hamiltonian in the orbital ground state is: ˆ2 , ˆ1 · S ˆ SS = A S H (8.1) h2 ¯ where the constant A has the dimension of an energy. ˆ SS in the spin basis {|σ1 , σ2 }, Recall the eigenstates and eigenvalues of H where σ1 = ±1, σ2 = ±1. 8.2.4. As in hydrogen, the constant A originates from a contact term: A=−

2 1 γ1 γ2 h ¯ 2 |ψ100 (0)|2 . 3 0 c2

(8.2)

(a) The observed hyperﬁne line of positronium has a frequency ν 200 GHz, compared to ν 1.4 GHz for hydrogen. Justify this diﬀerence of two orders of magnitude. (b) Express the constant A in terms of the ﬁne structure constant and the energy mc2 . Give the numerical value of A in eV. (c) What frequency of the hyperﬁne transition corresponds to this calculated value of A? 8.2.5. Actually, the possibility that the electron and the positron can annihiˆ A in the hyperﬁne Hamiltonian. One late, leads to an additional contribution H ˆ can show that HA does not aﬀect states of total spin equal to zero (S = 0), and that it increases systematically the energies of S = 1 states by the amount: ˆA : H

δE S=1 =

3A 4

(δE S=0 = 0) ,

(8.3)

where A is the same constant as in (8.2). (a) What are the energies of the S = 1 and S = 0 states, if one takes into account the above annihilation term? (b) Calculate the frequency of the corresponding hyperﬁne transition.

8.3 Zeeman Eﬀect in the Ground State The system is placed in a constant uniform magnetic ﬁeld B directed along the z axis. The additional Zeeman Hamiltonian has the form ˆ Z = ω1 Sˆ1z + ω2 Sˆ2z , H where ω1 = −γ1 B and ω2 = −γ2 B.

8.4 Decay of Positronium

75

8.3.1. (a) Taking into account the result of question 2.2 and setting ω = ˆ Z on the basis states {|σ1 , σ2 }. −γB, write the action of H (b) Write in terms of A and h ¯ ω the matrix representation of ˆA + H ˆZ ˆ =H ˆ SS + H H

(8.4)

in the basis {|S, m} of the total spin of the two particles. (c) Give the numerical value of ¯hω in eV for a ﬁeld B = 1 T. Is it easy experimentally to be in a strong ﬁeld regime, i.e. h ¯ ω A? 8.3.2. Calculate the energy eigenvalues in the presence of the ﬁeld B; express the corresponding eigenstates in the basis {|S, m} of the total spin. The largest eigenvalue will be written E+ and the corresponding eigenstate |ψ+ . For convenience, one can√introduce the quantity hω/(7A), and the angle √ x = 8¯ θ deﬁned by sin 2θ = x/ 1 + x2 , cos 2θ = 1/ 1 + x2 . 8.3.3. Draw qualitatively the variations of the energy levels in terms of B. Are there any remaining degeneracies?

8.4 Decay of Positronium We recall that when a system A is unstable and decays: A → B + · · · , the probability for this system to decay during the interval [t, t+dt] if it is prepared at t = 0, is dp = λe−λt dt, where the decay rate λ is related to the lifetime τ of the system by τ = 1/λ. If the decay can proceed via diﬀerent channels, e.g. A → B + · · · and A → C + · · · , with respective decay rates λ1 and λ2 , the total decay rate is the sum of the partial rates, and the lifetime of A is τ = 1/(λ1 + λ2 ). In all what follows, we place ourselves in the rest frame of the positronium. 8.4.1. In a two-photon decay, or annihilation, of positronium, what are the energies of the two outgoing photons, and what are their relative directions? 8.4.2. One can show that the annihilation rate of positronium into photons in an orbital state |n, l, m is proportional to the probability for the electron and positron to be at the same point, i.e. to |ψnlm (0)|2 . In what orbital states is the annihilation possible? 8.4.3. In quantum ﬁeld theory, one can show that, owing to charge conjugation invariance, (a) a singlet state, S = 0, can only decay into an even number of photons: 2, 4, · · · (b) a triplet state, S = 1, can only decay into an odd number of photons: 3, 5, · · · In the orbital ground state ψ100 , split by spin–spin interactions as calculated in Sect. 2, the lifetime of the singlet state is τ2 ∼ 1.25 × 10−10 s, and the

76

8 The Spectrum of Positronium

lifetime of either of the three triplet states is τ3 ∼ 1.4 × 10−7 s. Quantum ﬁeld theory predicts:

2 h ¯ 1 1 4 2 (π − 9) α λ2 . = 4πα2 c |ψ100 (0)|2 , λ3 = = λ2 = τ2 mc τ3 9π Compare theory and experiment. 8.4.4. In order to determine the hyperﬁne constant A of positronium, it is of interest to study the energy and the lifetime of the level corresponding to the state |ψ+ , deﬁned in question 3.2, as a function of the ﬁeld B. ¿From now on, we assume that the ﬁeld is weak, i.e. |x| = |8¯hω/(7A)| 1, and we shall make the corresponding approximations. (a) What are, as a function of x, the probabilities pS ans pT of ﬁnding the state |ψ+ in the singlet and triplet states respectively? + (b) Use the result to calculate the decay rates λ+ 2 and λ3 of the state |ψ+ into two and three photons respectively, in terms of the parameter x, and of the rates λ2 and λ3 introduced in question 4.4. (c) What is the lifetime τ + (B) of the state |ψ+ ? Explain qualitatively its dependence on the applied ﬁeld B, and calculate τ + (B) for B = 0.4 T . 1. One measures, as a function of B, the ratio R = τ + (B)/τ + (0) of the lifetime of the |ψ+ state with and without a magnetic ﬁeld. The dependence on B of R is given in Fig. 8.1, with the corresponding error bars. (i) What estimate does one obtain for the hyperﬁne constant, A, using the value of the magnetic ﬁeld for which the ratio R has decreased by a factor two? (ii) How do theory and experiment compare?

Fig. 8.1. Variation of the ratio R deﬁned in the text as a function of the applied magnetic ﬁeld B

8.5 Solutions

77

8.5 Solutions Section 8.1: Positronium Orbital States In positronium, we have, by scaling: 8.1.1. A reduced mass µ = m/2. ˆ =p ˆ 2 /2µ − q 2 /4π0 r. 8.1.2. A center of mass Hamiltonian H 8.1.3. The energy levels En = −(1/2)µc2 α2 /n2 = −(1/4)mc2 α2 /n2 . The degeneracy is n2 for each level, as in the hydrogen atom; the bound state energies are half of those of hydrogen. ˚ 8.1.4. The Bohr radius is a0 = h ¯ /(µcα) = 2¯ h/(mcα) = 2aH 0 ∼ 1.06 A. The , and, since the proton is ﬁxed, diameter of positronium is r = 3a0 /2 = 3aH 0 the diameter of the hydrogen is 2rH = 3aH . Therefore the two systems have 0 the same size. 8.1.5. The ground state wave function is ψ100 (r) = e−r/a0 / πa30 , and we 3 have |ψ100 (0)|2 = (mcα/(2¯ h)) /π. Section 8.2: Hyperﬁne Splitting 8.2.1. In the orbital ground state, the degeneracy is 4, corresponding to the number of independent spin states. 8.2.2. Since the masses are equal, but the charges are opposite, we have γ1 = q/m, γ2 = −q/m, γ = q/m. 8.2.3. As usual, we can express the spin–spin operator in terms of the total spin S as S 1 · S 2 = [S 2 − S12 − S22 ]/2. Hence, the orbital ground state is split into: √ (a) the triplet states: | + +, (| + − + | − +)/ 2, | − −, with the energy shift: E T = A/4 , √ (b) the singlet state: (| + − − | − +)/ 2, with the energy shift: E S = −3A/4 . 8.2.4. (a) There is a mass factor of ∼ 1/2000, a factor of ∼ 2.8 for the gyromagnetic ratio of the proton, and a factor of 8 due to the value of the wave function at the origin. Altogether, this results in a factor of ∼ 22/2000 ∼ 1% for the ratio of hyperﬁne splittings H/(e+ e− ). (b) The numerical value of A is A=

1 12π0

q¯ h mc

2

mcα 3 1 = mc2 α4 ∼ 4.84 × 10−4 eV. h ¯ 3

78

8 The Spectrum of Positronium

(c) This corresponds to a transition frequency of ν = A/h 117 GHz. This prediction is not in agreement with the experimental result (∼ 200 GHz). ˆ A , the triplet state energy is A while the 8.2.5. (a) Taking into account H singlet state energy is −3A/4. The splitting is δE = 7A/4 = 8.47 × 10−4 eV. (b) The corresponding frequency is ν = δE/h ∼ 205 GHz, in agreement with experiment. Section 8.3: Zeeman Eﬀect in the Ground State 8.3.1. ˆ Z = ω(Sˆ1z − Sˆ2z ), therefore, we have (a) The Zeeman Hamiltonian is H ˆ Z | + + = H ˆ Z | + − = H ˆ HZ | − + =

ˆ Z | − − = 0 H h ¯ ω| + − −¯ hω| − + .

In terms of total spin states, this results in ˆ Z |1, 1 = H ˆ Z |1, −1 = 0 H ˆ Z |1, 0 = h H ¯ ω|0, 0 ˆ Z |0, 0 = h H ¯ ω|1, 0 . (b) Hence the matrix representation in the coupled basis: ⎛

0 0 0 ⎜0 0 0 ˆZ = ⎜ H ⎝0 0 0 0 0 h ¯ω

⎞ 0 0 ⎟ ⎟ , ¯hω ⎠ 0

where the elements are ordered according to: |1, 1, |1, −1, |1, 0, |0, 0. 8.3.2. Similarly, one has the matrix tonian: ⎛ A 0 ⎜0 A ˆ =⎜ H ⎝0 0 0 0

representation of the full spin Hamil⎞ 0 0 0 0 ⎟ ⎟. A ¯hω ⎠ h ¯ ω −3A/4

In a ﬁeld of 1 T, |¯ hω| = q¯ hB/m = 2µB B 1.16 × 10−4 eV. The strong ﬁeld regime corresponds to |¯ hω| A , i.e. B 4 T, which is diﬃcult to reach.

8.5 Solutions

79

8.3.3. (a) Two eigenstates are obvious: |1, 1 and |1, −1, which correspond to the same degenerate eigenvalue A of the energy. The two others are obtained by diagonalizing a 2 × 2 matrix: |ψ+ = cos θ |1, 0 + sin θ |0, 0 , |ψ− = − sin θ |1, 0 + cos θ |0, 0 , corresponding to the energies A E± = ± 8

7A 8

1/2

2 2

+ (¯ hω)

=

A 1 ± 7 1 + x2 . 8

(b) The triplet states | + + and | − − remain degenerate, as shown in Fig. 8.2.

Fig. 8.2. Variation of the hyperﬁne energy levels with applied magnetic ﬁeld

Section 8.4: Decay of Positronium 8.4.1. In a two-photon decay, the outgoing photons have opposite momenta, their energies are both mc2 = 511 keV. 8.4.2. The wave function vanishes at the origin, except for s-waves (|ψnlm (0)|2 = 0 if l = 0), owing to the centrifugal barrier. Therefore the decay can only occur when the positronium is in an s-state. h) which yields τ2 = 8.4.3. The given formulas correspond to λ2 = mc2 α5 /(2¯ 1.24 × 10−10 s and τ3 = 1.38 × 10−7 s, in agreement with experiment.

80

8 The Spectrum of Positronium

8.4.4. (a) For a given value of the applied ﬁeld, with the positronium prepared in the state |ψ+ , the probabilities of ﬁnding the system in the singlet and triplet states are respectively pS = sin2 θ ∼ x2 /4 and pT = cos2 θ ∼ 1 − x2 /4. (b) The rate for |ψ+ to decay into two photons is the product of the probability of ﬁnding |ψ+ in the singlet state with the singlet state decay rate: S 2 2 λ+ 2 = p λ2 ∼ x λ2 /4 = x /(4τ2 ) .

Similarly, one has T 2 2 λ+ 3 = p λ3 ∼ (1 − x /4)λ3 = (1 − x /4)/τ3 .

(c) The lifetime of the |ψ+ state is τ+ =

λ+ 2

1 τ3 τ3 . + = 2 2 τ 2 2 + λ3 1 − x4 + x4 τ32 1 + 16¯h ω2 ττ32 49A

As the ﬁeld B increases, the state |ψ+ , which is purely triplet for B = 0, acquires a greater and greater singlet component. Therefore its lifetime decreases as B increases. For B = 0.4 T, one has τ+ = 0.23 τ3 = 3.2 × 10−8 s. (d) Experimentally, one has R ∼ 0.5, i.e. x2 τ3 /4τ2 1 for B ∼ 0.22 T. hω/7x and h ¯ ω = 2.3 × 10−5 eV, the Therefore x 6 × 10−2 and, since A = 8¯ −4 result is A ∼ 4.4 × 10 eV, in good agreement with theoretical expectations. Section 8.5: References S. DeBenedetti and H.C. Corben, Positronium, Ann. Rev. Nucl. Sci., 4, 191 (1954). Stephan Berko and Hugh N. Pendleton, Positronium, Ann. Rev. Nucl. Sci., 30, 543 (1980). A.P. Mills and S. Chu, Precision Measurements in Positronium, in Quantum Electrodynamics, ed. by T. Kinoshita (World Scientiﬁc, Singapore 1990) pp. 774-821.

9 The Hydrogen Atom in Crossed Fields

We study the modiﬁcation of the energy spectrum of a hydrogen atom placed in crossed static electric and magnetic ﬁelds in perturbation theory. We thus recover a result ﬁrst derived by Pauli. In his famous 1925 paper on the hydrogen atom, W. Pauli made use of the particular symmetry of the Coulomb problem. In addition to the hydrogen spectrum, he was able to calculate the splitting of the levels in an electric ﬁeld (Stark eﬀect) or in a magnetic ﬁeld (Zeeman eﬀect). Pauli also noticed that he could obtain a simple and compact formula for the level splitting in a superposition of a magnetic ﬁeld B 0 and an electric ﬁeld E 0 both static and uniform, and perpendicular to each other. In this case, he found that a level with principal quantum number n is split into 2n − 1 sublevels (k) En + δEn with ¯ k (ω02 + ωe2 )1/2 , (9.1) δEn(k) = h where k is an integer ranging from −(n−1) to n−1, ω0 and ωe are respectively proportional to B0 and E0 , and ωe can be written as ωe =

3 Ωe f (n) 2

with

Ωe =

4π0 ¯h E0 , M qe

where M and qe are the mass and charge of the electron, and where f (n) depends on n only. It is only in 1983 that Pauli’s result was veriﬁed experimentally. Our purpose, here, is to prove (9.1) in the special case n = 2, to calculate ω0 and ωe in that case, and, by examining the experimental result for n = 34, to guess what was the very simple formula found by Pauli for f (n).

82

9 The Hydrogen Atom in Crossed Fields

9.1 The Hydrogen Atom in Crossed Electric and Magnetic Fields We consider the n = 2 level of the hydrogen atom. We neglect all spin eﬀects. We assume that B 0 is along the z axis and E 0 along the x axis. We use ﬁrst order perturbation theory. 9.1.1. What are the energy levels and the corresponding eigenstates in the presence of B 0 only? Check that (9.1) is valid in this case and give the value of ω0 ? 9.1.2. In the presence of E 0 only, the perturbing Hamiltonian is the electric ˆ 0 = −qe rˆ .E 0 . Write the matrix representing H ˆ E in ˆ E = −D.E dipole term H the n = 2 subspace under consideration. We recall √ ∞ that: (a) 0 r3 R2s (r) R2p (r) dr = 3 3 a1 where R2s and R2p are the radial wave functions for the level n = 2, l = 0 and n = 2, l = 1 respectively, and ¯ 2 /(M e2 ) is the Bohr radius (e2 = qe2 /4π0 ). where a1 = h (b) In spherical coordinates (θ polar angle and φ azimuthal angle), the l = 0 and l = 1 spherical harmonics are 3 1 ±1 0 sin θ e±iφ , Y1 (θ, φ) = ∓ Y0 (θ, φ) = √ , 8π 4π 3 0 Y1 (θ, φ) = cos θ. (9.2) 4π 9.1.3. Calculate the energies of the levels originating from the n = 2 level in the presence of the crossed ﬁelds E 0 and B 0 . Show that one recovers (9.1) with ωe = (3/2)f (2)Ωe , and give the value of f (2).

9.2 Pauli’s Result The ﬁrst experimental veriﬁcation of Pauli’s result was performed in 1983.1 In Fig. 9.1, the points correspond to a sub-level with a given value of k arising from the n = 34 level of an hydrogen-like atom. All points correspond to the same energy of this level, but to diﬀerent values of the static ﬁelds E0 and B0 . Knowing that ωe is a function of the principal quantum number n of the form: ωe = (3/2)f (n)Ωe , and that ω0 and Ωe are the constants introduced above, answer the following questions: 1

Fig. 9.1 was obtained by F. Biraben, D. Delande, J.-C. Gay, and F. Penent, with rubidium atoms prepared in a Rydberg state, i.e. with an electron placed in a strongly excited level (see J.-C. Gay, in Atoms in unusual situations, J.-P. Briand ed., p. 107, Plenum, New York, 1986).

9.3 Solutions

83

Fig. 9.1. Values of the electric and magnetic ﬁelds giving rise to the same sub-level energy of the n = 34 level of a hydrogen-like atom

9.2.1. Does the experimental data agree with (9.1)?

9.2.2. Write the quantity ω02 + ωe2 in the form λ γB02 + f 2 (n)E02 , give the value of the constant γ, and calculate f (34). 9.2.3. Guess Pauli’s result concerning f (n).

9.3 Solutions Section 9.1: The Hydrogen Atom in Crossed Electric and Magnetic Fields 9.1.1. Consider a state |n, l, m. The orbital magnetic moment of the elecˆ with γ0 = qe /(2M ). The magnetic Hamiltonian is ˆ orb = γ0 L, tron is µ ˆ ˆ z B0 . ˆ orb .B = −(qe /2M )L H = −µ At ﬁrst order perturbation theory, the energy levels originating from the n = 2 subspace (angular momentum l = 0 or l = 1) are m¯hω0 with m = −1, 0, +1, and ω0 = −qe B0 /(2M ) (ω0 > 0 for B0 > 0). The corresponding states are |2s and |2p, m = 0 |2p, m = −1 |2p, m = +1

δE = 0 δE = −¯ hω0 δE = +¯ hω0 .

ˆ E = −qe x 9.1.2. The Hamiltonian is H ˆ E0 . We have to calculate the 16 matrix x|2, l, m. The integral to be evaluated is elements 2, l , m |ˆ

84

9 The Hydrogen Atom in Crossed Fields

2, l , m |ˆ x|2, l, m =

×

∞

∗ Ylm (θ, φ) sin θ cos φ Ylm (θ, φ) d2 Ω ∗

r3 (R2,l (r)) R2,l (r) dr .

0

The angular integral vanishes if l = l . We need only consider the terms l = 0, l = 1 (and the hermitian conjugate l = 1, l = 0), i.e. √ 2π 1 √ 3 3 a1 (−Y11 (θ, φ) + Y1−1 (θ, φ)) Y1m (θ, φ) d2 Ω 3 4π where we have incorporated √ the radial integral given in the text. One therefore obtains 3 a1 (δm,−1 − δm,1 )/ 2. The only non-vanishing matrix elements are ˆ 2s|H|2p, m = ±1 and their hermitian conjugates. hE0 /(M qe ) = qe E0 a1 /¯ h, we obtain the matrix Setting Ωe = 4π0 ¯ ⎛ ⎞ 0 0 0 1 hΩ ⎜ 0 0 0 0⎟ ˆ E = 3¯ ⎟ . √ e⎜ H 0 −1 ⎠ 2 ⎝0 0 1 0 −1 0 where the rows (columns) are ordered as 2p, m = 1, 0, −1; 2s. 9.1.3. We want to ﬁnd the eigenvalues of the matrix √ ⎞ ⎛ ω0 0 0 3Ωe / 2 ⎜ 0 0 0 0 √ ⎟ ⎟ . h⎜ ¯ ⎝ 0√ 0 −ω0√ −3Ωe / 2 ⎠ 3Ωe / 2 0 −3Ωe / 2 0 There is an obvious eigenvalue λ = 0 since the |2p, m = 0 and |2s states do not mix in the presence the electric ﬁeld. The three other eigenvalues are easily obtained as the solutions of: λ(¯ h2 ω02 − λ2 ) + 9 h ¯ 2 Ωe2 λ = 0 , i.e. λ = 0 and λ = ±¯ h ω02 + 9Ωe2 . The shifts of the energy levels are therefore: δE = 0 twice degenerate, and δE = ±¯ h ω02 + 9Ωe2 . If we adopt the prescription given in the text, we obtain ωe = 3Ωe =⇒ f (2) = 2 . Section 9.2: Pauli’s Result 9.2.1. We remark that the experimental points are aligned on a straight line aB02 + bE02 = constant which is in agreement with (9.1), i.e. a constant value of ω02 + ωe2 corresponds to a constant value of each energy level.

9.3 Solutions

85

9.2.2. Given the deﬁnitions of ω0 and Ωe , one has ω02 + ωe2 =

9 4

h 4π0 ¯ M qe

2

αc 2 2 B0 + f 2 (n) E02 3

,

where α is the ﬁne structure constant and c the velocity of light. The experimental line 2 (αc/3) B02 + f 2 (34)E02 goes through the points E02 = 0, B02 87 × 10−4 T2 and B02 = 0, E02 4 × 106 V2 m−2 . This gives f (34) = 34. 9.2.3. Indeed, the very simple result found by Pauli was f (n) = n.

10 Energy Loss of Ions in Matter

When a charged particle travels through condensed matter, it loses its kinetic energy gradually by transferring it to the electrons of the medium. In this chapter we evaluate the energy loss of the particle as a function of its mass and its charge, by studying the modiﬁcations that the state of an atom undergoes when a charged particle passes in its vicinity. We show how this process can be used to identify the products of a nuclear reaction. The electric potential created by the moving particle appears as a timedependent perturbation in the atom’s Hamiltonian. In order to simplify the problem, we shall consider the case of an atom with a single external electron. The nucleus and the internal electrons will be treated globally as a core of charge +q, inﬁnitely massive and, therefore, ﬁxed in space. We also assume that the incident particle of charge Z1 q is heavy and non-relativistic, and that its kinetic energy is large enough so that in good approximation its motion can be considered linear and uniform, of constant velocity v, when it interacts with an atom. Here q denotes the unit charge and we set e2 = q 2 /(4πε0 ). We consider the x, y plane deﬁned by the trajectory of the particle and the center of gravity of the atom, which is chosen to be the origin, as shown on Fig. 10.1. Let R(t) be the position of the particle at time t and r = (x, y, z) the coordinates of the electron of the atom. The impact parameter is b and the notation is speciﬁed in Fig. 10.1. The time at which the particle passes nearest to the atom, i.e. x = b, y = 0 is denoted t = 0. We write En and |n for the energy levels and corresponding eigenstates of the atom in the absence of an external perturbation.

10.1 Energy Absorbed by One Atom 10.1.1. Write the expression for the time-dependent perturbing potential Vˆ (t) due to the presence of the charged particle.

88

10 Energy Loss of Ions in Matter

10.1.2. We assume that the impact parameter b is much larger than the typical atomic size, i.e. b r, so that |R(t)| |r| for all t. Replace Vˆ (t) by its ﬁrst order expansion in |r|/|R| and express the result in terms of the coordinates x and y of the electron, and of b, v and t. 10.1.3. Initially, at time t = −∞, the atom is in a state |i of energy Ei . Using ﬁrst order time-dependent perturbation theory, write the probability amplitude γif to ﬁnd the atom in the ﬁnal state |f of energy Ef after the h and we only charged particle has passed (t = +∞). We set ωf i = (Ef − Ei )/¯ consider the case Ef = Ei . 10.1.4. The calculation of γif involves the Bessel function K0 (z). One has ∞ ∞ cos ωt sin ωt dt = K0 (ωβ) t 2 dt = ω K0 (ωβ) . 2 + t2 )1/2 (β (β + t2 )3/2 0 0 Express γif in terms of K0 and its derivative. behavior of K0 is K0 (z) − ln z for z 1, and K0 (z) The asymptotic 2π/z e−z for z 1. Under what condition on the parameters ωf i , b and v is the transition probability Pif = |γif |2 large? Show that, in that case, one obtains

Pif

2Z1 e2 hbv ¯

2 |f |ˆ x|i|2 .

10.1.5. Give the physical interpretation of the condition derived above. Show that, given the parameters of the atom, the crucial parameter is the eﬀective interaction time, and give a simple explanation of this eﬀect.

10.2 Energy Loss in Matter We assume in the following that the Hamiltonian of the atom is of the form

Fig. 10.1. Deﬁnition of the coordinates.

10.2 Energy Loss in Matter

89

ˆ2 ˆ0 = p + V (ˆ r) . H 2m 10.2.1. Thomas–Reiche–Kuhn Sum Rule. ˆ 0 ]. (a) Calculate the commutator [ˆ x, H (b) Deduce from this commutator a relation between the matrix elements ˆ 0. i|ˆ x|f and i|ˆ p|f , where |i and |f are eigenstates of H (c) Applying a closure relation to [ˆ x, pˆ] = i¯ h, show that: 2m (Ef − Ei )|f |ˆ x|i|2 = 1 h2 f ¯ for all eigenstates |i of H0 . 10.2.2. Using the Thomas–Reiche–Kuhn sum rule, calculate the expectation value δE of the energy loss of the incident particle when it interacts with the atom. Let E be the energy of the particle before the interaction. Which parameters does the product E δE depend on? 10.2.3. Experimental Application. We are now interested in incident particles which are fully ionized atoms (Z1 = Z, where Z is the atomic number), whose masses are, to a good approximation, proportional to the mass number A = Z + N (where N is the number of neutrons of the isotope). When these ions traverse condensed matter, they interact with many atoms of the medium, and their energy loss implies some averaging over the random impact parameter b. The previous result then takes the form E δE = kZ 2 A , where the constant k depends on the nature of the medium. Semiconductor detectors used for the identiﬁcation of the nuclei in nuclear reactions are based on this result. In the following example, the ions to be identiﬁed are the ﬁnal state products of a reaction induced by 113 MeV nitrogen ions impinging on a target of silver atoms. In Fig. 10.2 each point represents an event, i.e. the energy E and energy loss δE of an ion when it crosses a silicon detector. The reference point corresponds A to the isotope A = 12 of carbon 12 6 C (we use the notation Z N for a nucleus charge Z and mass number A) which loses δE = 30 MeV at an energy E = 50 MeV. (a) Calculate the constant k and the theoretical prediction for the energy loss at 60 and 70 MeV. Put the corresponding points on the ﬁgure. (b) Assuming the reaction could produce the following isotopes: – boron, Z = 5, A = 10, 11, 12 – carbon, Z = 6, A = 11, 12, 13, 14 – nitrogen Z = 7, A = 13, 14, 15, 16,

90

10 Energy Loss of Ions in Matter 10

20

30

40

50

60

70

80

90

100

E(MeV)

10

20

30 12 6

C

40 14 7

N(113MeV)→Ag

50

δE(MeV) Fig. 10.2. Energy loss δE versus energy E through a silicon detector, of the ﬁnal products of a reaction corresponding to 113 MeV nitrogen ions impinging on a target of silver atoms

what nuclei are eﬀectively produced in the reaction? Justify your answers by putting the points corresponding to E = 50 MeV and E = 70 MeV on the ﬁgure.

10.3 Solutions Section 10.1: Energy Absorbed by One Atom 10.1.1. The interaction potential between the particle and the atom is the sum of the Coulomb interactions between the particle and the core, and those between the particle and the outer electron: Z1 e2 Z1 e2 − . Vˆ (t) = R(t) |R(t) − rˆ | 10.1.2. For |R| |r|, we have

−1/2 1 r·R 1 = R2 − 2R · r + r2 + . |R − r| R R3

10.3 Solutions

Therefore

91

2

Z1 e Vˆ (t) − 3 rˆ · R(t) . R (t) Since R(t) = (b, vt, 0),we obtain Vˆ (t) −

Z1 e2 (b2 + v 2 t2 )

3/2

(ˆ xb + yˆvt) .

10.1.3. To ﬁrst order in Vˆ , the probability amplitude is 1 +∞ iωf i t ˆ e f |V (t)|i dt . γif = i¯ h −∞ Inserting the value found above for Vˆ (t), we ﬁnd 1 +∞ Z1 e2 eiωf i t (bf |ˆ x|i + vtf |ˆ y |i) dt . γif = − i¯ h −∞ (b2 + v 2 t2 )3/2 10.1.4. One has ∞ 0

cos ωt dt (β 2

+

3/2 t2 )

=−

1 d ω K0 (ωβ) = − K0 (ωβ) . β dβ β

Setting β = b/v, the amplitude γif is γif = i

2Z1 e2 ωf i (K0 (ωf i b/v) f |ˆ y |i − K0 (ωf i b/v) f |ˆ x|i) . hv 2 ¯

The probability Pif = |γif |2 is large if K0 or K0 are also large. This happens for ωf i b/v 1. In this limit, K0 (z) ∼ − ln z and K0 (z) ∼ −1/z, and we obtain

ωf i b 2Z1 e2 ωf i b ln γif = i f |ˆ x|i − f |ˆ y |i . hvb ¯ v v Since |f |ˆ x|i| |f |ˆ y |i|, one can neglect the second term (x ln x 1 for x 1) and we obtain, for ωf i b/v 1,

Pif = |γif | 2

2Z1 e2 hbv ¯

2 |f |ˆ x|i|2 .

10.1.5. The time τ = b/v is the characteristic time during which the interaction is important, as we can see on the above formulas. For t τ , the interaction is negligible. The condition ωf i τ 1 means that the interaction time τ must be much smaller than the Bohr period ∼ 1/ωf i of the atom. The perturbation Vˆ (t) must have a large Fourier component at ω = ωf i if we want the probability Pif to be signiﬁcant (the shorter in time the perturbation, the larger the

92

10 Energy Loss of Ions in Matter

spread of its Fourier transform in frequency). In the opposite limiting case, where the perturbation is inﬁnitely slow, the atom is not excited. This observation provides an alternative way to evaluate the integrals of question 1.3. The only values of t which contribute signiﬁcantly are those for which t is not too large, compared to τ (say |t| 10 τ ). If ωf i τ 1, one can replace eiωf i t by 1 in these integrals; the second integral is then zero for symmetry reasons and the ﬁrst one is easily evaluated, and gives the desired result. Section 10.2: Energy Loss in Matter 10.2.1. Thomas–Reiche–Kuhn Sum Rule. ˆ 0 ] = i¯ hpˆ/m. (a) We ﬁnd [ˆ x, H (b) Taking the matrix element of this commutator between two eigenstates ˆ 0 , we obtain: |i and |f of H i¯ h ˆ 0 ] | i = (Ei − Ef )f | x f | pˆ | i = f | [ˆ x, H ˆ | i . m (c) We now take the matrix element of [ˆ x, pˆ] = i¯ h between i| and |i and we use the closure relation: i¯ h=

i | x ˆ | f f | pˆ | i −

f

i | pˆ | f f | x ˆ | i

f

m (Ei − Ef ) | f | x ˆ | i | i¯ h

2

2m = (Ei − Ef )| f | x ˆ | i | i¯ h

2

=

−

f

m (Ef − Ei ) | i | x ˆ | f | i¯ h

2

f

,

f

which proves the Thomas–Reiche–Kuhn sum rule. 10.2.2. The expectation value δE of the energy transferred to the atom is δE =

f

(Ef − Ei ) Pif =

2Z1 e2 hbv ¯

2

(Ef − Ei ) | f | x ˆ | i |

2

.

f

Making use of the Thomas–Reiche–Kuhn sum rule, we obtain δE =

2Z12 e4 , mb2 v 2

where m is the electron mass. If the ion has mass M , its kinetic energy is E = M v 2 /2, and we therefore obtain a very simple expression:

10.3 Solutions

M E δE = m

Z1 e2 b

93

2 ,

where we see that the product E δE does not depend on the energy of the incident particle, but is proportional to its mass and to the square of its charge. 10.2.3. With the 126 C point, one obtains k = 3.47. We have put the calculated points of the various isotopes on Fig. 10.3. We make the following observations: (a) For boron, the isotopes 10 B and 11 B are produced, but not 12 B. (b) For carbon, 12 C is produced more abundantly than 13 C, 14 C and 11 C. (c) For nitrogen, there is an abundant production of 14 N, a smaller production of 15 N, but practically no 13 N or 16 N. 10

20

30

10

40

50

60

70

80

10 11 5

90

B

100

11 12 13 14

6

E(MeV)

C

20 14 15 7

N

30 12 6

C

40 14 7

N(113MeV)→Ag

50

δE(MeV) Fig. 10.3. Interpretation of the data of Fig. 10.2

94

10 Energy Loss of Ions in Matter

10.4 Comments Ionization of matter has numerous applications, for instance in developing detectors for particle and nuclear physics, or in deﬁning protection regulations against radioactivity. In order to calculate the energy loss of an ion in matter, one must integrate the above results over the impact parameter. In practice, taking everything into account, one ends up with the following formula, due to Hans Bethe and Felix Bloch, for the rate of energy loss per unit length:

2me c2 β 2 4πK 2 Z 2 e4 N dE = (ln (10.1) − β2) − dx me c2 β 2 I(1 − β 2 ) where β = v/c, K is a constant, N is the number density of atoms in the medium and I is the mean excitation energy of the medium (I ∼ 11.5 eV). The cases of protons or heavy ions is of great interest and, in comparatively recent years, it has allowed a major improvement in the medical treatment of tumors in the eyes (proton therapy) and in the brain (ion therapy) . Owing to the factor 1/β 2 , or equivalently 1/v 2 , in (10.1), practically all the energy is deposited in a very localized region near the stopping point. Figure (10.4) shows the comparison between the eﬀect of ion beams and photons. One can see the enormous advantage, from the medical point of view, of heavy ion beams. These permit to attack and destroy tumors in a very accurate and localized manner, as opposed to γ rays which produce damages all around the point of interest. Pioneering work on brain tumor therapy has been developed in Darmstadt at the Heavy ion accelerator facility. Information can be found on the sites

Fig. 10.4. Energy loss of ions (left) and survival rate of cells (right) as a function of the penetration depth. The dashed curve corresponds to the same quantities for photons. We can see the considerable medical advantage to use heavy ion beams. Document from the data of Heavy ion therapy at GSI, Darmstadt, http://www.gsi.de (Courtesy James Rich)

10.4 Comments

95

http://www-aix.gsi.de/ bio/home.html http://www.sgsmp.ch/protsr-f.htm This promising sector of medical applications in rapidly developing at present.

Part II

Quantum Entanglement and Measurement

12 Schr¨ odinger’s Cat

The superposition principle states that if |φa and |φb are two possible states √ of a quantum system, the quantum superposition (|φa + |φb )/ 2 is also an allowed state for this system. This principle is essential in explaining interference phenomena. However, when it is applied to “large” objects, it leads to paradoxical situations where a system can be in a superposition of states which is classically self-contradictory (antinomic). The most famous example is Schr¨odinger’s “cat paradox” where the cat is in a superposition of the “dead” and “alive” states. The purpose of this chapter is to show that such superpositions of macroscopic states are not detectable in practice. They are extremely fragile, and a very weak coupling to the environment suﬃces to destroy the quantum superposition of the two states |φa and |φb .

12.1 The Quasi-Classical States of a Harmonic Oscillator In this chapter, we shall consider high energy excitations of a one-dimensional harmonic oscillator, of mass m and frequency ω. The Hamiltonian is written 2 ˆ = pˆ + 1 mω 2 x ˆ2 . H 2m 2

ˆ by {|n}. The energy of the state |n is En = We denote the eigenstates of H (n + 1/2)¯ hω. ˆ = x mω/¯h, Pˆ = 12.1.1. Preliminaries. We introduce the operators X ˆ √ pˆ/ m¯hω and the annihilation and creation operators 1 ˆ + iPˆ a ˆ= √ X 2

1 ˆ a ˆ† = √ X − iPˆ 2

ˆ =a N ˆ† a ˆ.

110

12 Schr¨ odinger’s Cat

ˆ Pˆ ] = i, [ˆ ˆ = We recall the commutators: [X, a, a ˆ† ] = 1, and the relations: H ˆ ˆ hω(N + 1/2) and N |n = n|n. ¯ (a) Check that if one works with functions of the dimensionless variables X and P , one has ∂ ˆ =i ∂ . X Pˆ = −i ∂X ∂P ˆ (b) Evaluate the commutator [N , a ˆ], and prove that a ˆ|n =

√

n|n − 1

(12.1)

up to a phase factor which we set equal to 1 in what follows. ˆ and Pˆ , calculate (c) Using (12.1) for n = 0 and expressing a ˆ in terms of X the wave function of the ground state ψ0 (X) and its Fourier transform ϕ0 (P ). It is not necessary to normalize the result. 12.1.2. The Quasi-Classical States. The eigenstates of the operator a ˆ are called quasi-classical states, for reasons which we now examine. Consider an arbitrary complex number α. Show that the following state |α = e−|α|

2

/2

αn √ |n n! n

(12.2)

is a normalized eigenstate of a ˆ with eigenvalue α: a ˆ|α = α|α. 12.1.3. Calculate the expectation value of the energy in a quasi-classical state |α. Calculate also the expectation values x and p and the root mean square deviations ∆x and ∆p for this state. Show that one has ∆x ∆p = h ¯ /2. 12.1.4. Following a similar procedure as in question 1.1(c) above, determine the wave function ψα (X) of the quasi-classical state |α, and its Fourier transform ϕα (P ). Again, it is not necessary to normalize the result. 12.1.5. Suppose that at time t = 0, the oscillator is in a quasi-classical state |α0 with α0 = ρeiφ where ρ is a real positive number. (a) Show that at any later time t the oscillator is also in a quasi-classical state which can be written as e−iωt/2 |α(t). Determine the value of α(t) in terms of ρ, φ, ω and t. (b) Evaluate xt and pt . Taking 1.3, and assuming that |α| 1, justify brieﬂy why these states are called “quasi-classical”. 12.1.6. Numerical Example. Consider a simple pendulum of length 1 meter and of mass 1 gram. Assume the state of this pendulum can be described by a quasi-classical state. At time t = 0 the pendulum is at x0 = 1 micrometer from its classical equilibrium position, with zero mean velocity. (a) What is the corresponding value of α(0)? (b) What is the relative uncertainty on its position ∆x/x0 ? (c) What is the value of α(t) after 1/4 period of oscillation?

12.3 Quantum Superposition Versus Statistical Mixture

111

12.2 Construction of a Schr¨ odinger-Cat State During the time interval [0, T ], one adds to the harmonic potential, the coupling ˆ =h W ¯ g (ˆ a† a ˆ)2 . We assume that g is much larger than ω and that ωT 1. Hence, we can make the approximation that, during the interval [0, T ], the Hamiltonian of ˆ . At time t = 0, the system is in a quasi-classical state the system is simply W |ψ(0) = |α. ˆ , and write the expan12.2.1. Show that the states |n are eigenstates of W sion of the state |ψ(T ) at time T on the basis {|n}. 12.2.2. How does |ψ(T ) simplify in the particular cases T = 2π/g and T = π/g? 12.2.3. One now chooses T = π/2g. Show that this gives 1 −iπ/4 |α + eiπ/4 | − α . e |ψ(T ) = √ 2

(12.3)

12.2.4. Suppose α is pure imaginary: α = iρ. (a) Discuss qualitatively the physical properties of the state (12.3). (b) Consider a value of |α| of the same order of magnitude as in 1.6. In what sense can this state be considered a concrete example of the “Schr¨odinger cat” type of state mentioned in the introduction?

12.3 Quantum Superposition Versus Statistical Mixture We now study the properties of the state (12.3) in a “macroscopic” √ situation hω. |α| 1. We choose α pure imaginary, α = iρ, and we set p0 = ρ 2m¯ 12.3.1. Consider a quantum system in the state (12.3). Write the (non-normalized) probability distributions for the position and for the momentum of the system. These probability distributions are represented in Fig. 12.1 for α = 5i. Interpret these distributions physically. 12.3.2. A physicist (Alice) prepares N independent systems all in the state (12.3) and measures the momentum of each of these systems. The measuring apparatus has a resolution δp such that: √ m¯hω δp p0 . For N 1, draw qualitatively the histogram of the results of the N measurements.

112

12 Schr¨ odinger’s Cat

Fig. 12.1. Probability distributions for the position and for the momentum of a system in the state (12.3) for α = 5i. The quantities X and P are the dimensionless variables introduced in the ﬁrst part of the problem. The vertical scale is arbitrary

12.3.3. The state (12.3) represents the quantum superposition of two states which are macroscopically diﬀerent, and therefore leads to the paradoxical situations mentioned in the introduction. Another physicist (Bob) claims that the measurements done by Alice have not been performed on N quantum systems in the state (12.3), but that Alice is actually dealing with a nonparadoxical “statistical mixture”, that is to say that half of the N systems are in the state |α and the other half in the state | − α. Assuming this is true, does one obtain the same probability distribution as for the previous question for the N momentum measurements? 12.3.4. In order to settle the matter, Alice now measures the position of each of N independent systems, all prepared in the state (12.3). Draw the shape of the resulting distribution of events, assuming that the resolution δx of the measuring apparatus is such that: h ¯ 1 . δx |α| mω 12.3.5. Can Bob obtain the same result concerning the N position measurements assuming he is dealing with a statistical mixture? 12.3.6. Considering the numerical value obtained in the case of a simple pendulum in question 1.6, evaluate the resolution δx which is necessary in order to tell the diﬀerence between a set of N systems in the quantum superposition (12.3), and a statistical mixture consisting in N/2 pendulums in the state |α and N/2 pendulums in the state | − α.

12.4 The Fragility of a Quantum Superposition In a realistic physical situation, one must take into account the coupling of the oscillator with its environment, in order to estimate how long one can discriminate between the quantum superposition (12.3) (that is to say the “Schr¨ odinger cat” which is “alive and dead”) and a simple statistical mixture (i.e. a set of cats (systems), half of which are alive, the other half being dead; each cat being either alive or dead.)

12.4 The Fragility of a Quantum Superposition

113

If the oscillator is initially in the quasi-classical state |α0 and if the environment is in a state |χe (0), the wave function of the total system is the product of the individual wave functions, and the state vector of the total system can be written as the (tensor) product of the state vectors of the two subsystems: |Φ(0) = |α0 |χe (0) . The coupling is responsible for the damping of the oscillator’s amplitude. At a later time t, the state vector of the total system becomes: |Φ(t) = |α1 |χe (t) with α1 = α(t)e−γt ; the number α(t) corresponds to the quasi-classical state one would ﬁnd in the absence of damping (question 1.5(a)) and γ is a real positive number. 12.4.1. Using the result 1.3, give the expectation value of the energy of the oscillator at time t, and the energy acquired by the environment when 2γt 1. 12.4.2. For initial states of the “ Schr¨ odinger cat” type for the oscillator, the state vector of the total system is, at t = 0, 1 |Φ(0) = √ e−iπ/4 |α0 + eiπ/4 | − α0 |χe (0) 2 and, at a later time t, 1 iπ/4 | − α1 |χ(−) |Φ(t) = √ e−iπ/4 |α1 |χ(+) e (t) + e e (t) 2 still with α1 = α(t)e−γt . We choose t such that α1 is pure imaginary, with (+) (−) |α1 | 1. |χe (t) and |χe (t) are two normalized states of the environment that are a priori diﬀerent (but not orthogonal). The probability distribution of the oscillator’s position, measured independently of the state of the environment, is then P(x) =

(+)

1 [|ψα1 (x)|2 + |ψ−α1 (x)|2 2 (−) + 2Re(iψα∗ 1 (x)ψ−α1 (x)χ(+) e (t)|χe (t))]. (−)

Setting η = χe (t)|χe (t) with 0 ≤ η ≤ 1 (η is supposed to be real) and using the results of Sect. 3, describe without any calculation, the result of: (a) N independent position measurements, (b) N independent momentum measurements. Which condition on η allows one to distinguish between a quantum superposition and a statistical mixture?

114

12 Schr¨ odinger’s Cat

12.4.3. In a very simple model, the environment is represented by a second oscillator, of same mass and frequency as the ﬁrst one. We assume that this second oscillator is initially in its ground state |χe (0) = |0. If the coupling between the two oscillators is quadratic, we will take for granted that (±)

(±)

• the states |χe (t) are quasi-classical states: |χe (t) = | ± β, • and that, for short times (γt 1): |β|2 = 2γt|α0 |2 . (a) From the expansion (12.2), show that η = β| − β = exp(−2|β|2 ). (b) Using the expression found in question 4.1 for the energy of the ﬁrst oscillator, determine the typical energy transfer between the two oscillators, above which the diﬀerence between a quantum superposition and a statistical mixture becomes unobservable. 12.4.4. Consider again the simple pendulum described above. Assume the damping time is one year (a pendulum in vacuum with reduced friction). Using the result of the previous question, evaluate the time during which a “Schr¨ odinger cat” state can be observed. Comment and conclude.

12.5 Solutions Section 12.1: The Quasi-Classical States of a Harmonic Oscillator 12.1.1. (a) A simple change of variables gives ∂ p ˆ ∂ 1 ¯h ∂ h ¯ = −i = −i Pˆ = √ =√ mω ∂x ∂X m¯hω m¯hω i ∂x √ mω ∂ ∂ ∂ ˆ = mω x X ˆ= i¯ h = i m¯hω =i h ¯ h ¯ ∂p ∂p ∂P ˆ, a (b) We have the usual relations [N ˆ] = [ˆ a† a ˆ, a ˆ] = [ˆ a† , a ˆ]ˆ a = −ˆ a. Consequently: ˆ, a ˆa [N ˆ]|n = −ˆ a|n ⇒ N ˆ|n = (n − 1)ˆ a|n , ˆ corresponding to the eigenvalue n−1. We know and a ˆ|n is an eigenvector of N from the theory of the one-dimensional harmonic oscillator that the energy levels are not degenerated. Therefore we ﬁnd that a ˆ|n = µ|n − 1, where the coeﬃcient µ is determined by calculating the norm of a ˆ|n: √ a† a ˆ|n = n ⇒ µ = n ˆ a|n2 = n|ˆ up to an arbitrary phase. ˆ + iPˆ )|0 = 0. (c) The equation a ˆ|0 = 0 corresponds to (X

∂ In real space: X + ∂X ψ0 (X) = 0 ⇒ ψ0 (X) ∝ exp −X 2 /2 .

∂ ϕ0 (P ) = 0 ⇒ ϕ0 (P ) ∝ exp −P 2 /2 . In momentum space: P + ∂P

12.5 Solutions

115

12.1.2. One can check directly the relation a ˆ|α = α|α: αn αn √ 2 √ a √ ˆ |n = e−|α| /2 n |n − 1 n! n! n n αn 2 √ |n = α|α = αe−|α| /2 n! n

a ˆ|α = e−|α|

2

/2

The calculation of the norm of |α yields: α|α = e−|α|

2

n

|α|2n n!

=1.

12.1.3. The expectation value of the energy is: ˆ ˆ + 1/2|α = h E = α|H|α =h ¯ ωα|N ¯ ω(|α|2 + 1/2) . For x, and p, we use ¯h h ¯ α|ˆ a+a ˆ† |α = (α + α∗ ) x = 2mω 2mω m¯hω mω¯ h ∗ † α|ˆ a−a ˆ |α = i (α − α) p = −i 2 2 h ¯ h ¯ α|(ˆ a+a ˆ† )2 |α − x2 = ((α + α∗ )2 + 1) − x2 . ∆x2 = 2mω 2mω h/2mω, which is independent of α. Therefore ∆x = ¯ Similarly m¯hω m¯hω α|(ˆ a−a ˆ† )2 |α − p2 = − ((α − α∗ )2 − 1) − p2 2 2 Therefore ∆p = m¯hω/2. The Heisenberg inequality becomes in this case an equality ∆x ∆p = h ¯ /2, independently of the value of α. ∆p2 = −

12.1.4. With the X variable, we have

1 ∂ √ X+ ψα (X) = αψα (X) ∂X 2

√ (X − α 2)2 ⇒ ψα (X) = C exp − 2

Similarly, with the P variable,

i ∂ √ P+ ϕα (P ) = αϕα (P ) ∂P 2

√ (P + iα 2)2 ⇒ ϕα (P ) = C exp − . 2

116

12 Schr¨ odinger’s Cat

12.1.5. (a) |ψ(0) = |α0

αn √ 0 e−iEn t/¯h |n n! n αn 2 √ 0 e−inωt |n = e−|α| /2 e−iωt/2 n! n

|ψ(t) = e−|α|

2

/2

= e−iωt/2 |α(t)

with α(t) = α0 e−iωt = ρe−i(ωt−φ)

(b) xt = 2¯h/(mω) ρ cos(ωt − φ) = x0√cos(ωt − φ) pt = − 2m¯ hω ρ sin(ωt − φ) = −p0 sin(ωt − φ)

with

x0 = ρ

2¯h/(mω)

with

√ p0 = ρ 2m¯ hω

These are the equations of motions of a classical oscillator. Using the result 1.3, we obtain ∆p 1 1 ∆x 1 , 1 = = x0 2ρ p0 2ρ The relative uncertainties on the position and on the momentum of the oscillator are quite accurately deﬁned at any time. Hence the name “quasi-classical state”. 12.1.6. (a) The appropriate choice is x0 = x0 and p0 = 0, i.e. φ = 0 g = 3.13 s−1 ⇒ α(0) = 3.9 109 ω = 2πν = (b) ∆x/x0 = 1/(2α(0)) = 1.3 10−10 . (c) After 1/4 period, eiωt = eiπ/2 = i ⇒ α(T /4) = −i 3.9 109 Section 12.2: Construction of a Schr¨ odinger-Cat State ˆ are simply the previous |n, therefore: 12.2.1. The eigenvectors of W ˆ |n = h W ¯ g n2 |n and |ψ(0) = |α

⇒

|ψ(T ) = e−|α|

12.2.2. If T = 2π/g, then e−ign

2

T

2

2

/2

αn 2 √ e−ign T |n n! n

= e−2iπn = 1 and

|ψ(T ) = |α .

.

12.5 Solutions

If T = π/g, then e−ign

2

T

e−ign

2

2

= e−iπn = 1 if n is even, −1 if n is odd, therefore T

= (−1)n

12.2.3. If T = π/2g, then e−ign if n is odd. We can rewrite this relation as e−ign

2

T

=

117

2

T

⇒

|ψ(T ) = | − α π

2

= e−i 2 n = 1 for n even, and e−ign

2

T

= −i

π π 1 1 (1 − i + (1 + i)(−1)n ) = √ (e−i 4 + ei 4 (−1)n ) 2 2

or, equivalently, 1 |ψ(T ) = √ (e−iπ/4 |α + eiπ/4 | − α) . 2 12.2.4. (a) For α = iρ, in the state |α, the oscillator has a zero mean position and a positive velocity. In the state | − α, the oscillator also has a zero mean position, but a negative velocity. The state 12.3 is a quantum superposition of these two situations. (b) If |α| 1, the states |α and | − α are macroscopically diﬀerent (antinomic). The state 12.3 is a quantum superposition of such states. It therefore constitutes a (peaceful) version of Schr¨ odinger’s cat, where we represent “dead” or “alive” cats by simple vectors of Hilbert space. Section 12.3: Quantum Superposition Versus Statistical Mixture 12.3.1. The probability distributions of the position and of the momentum are P (X) ∝ |e−iπ/4 ψα (X) + eiπ/4 ψ−α (X)|2

√ 2 −iπ/4 1 ∝ e exp − (X − iρ 2) 2

2 √ 1 +eiπ/4 exp − (X + iρ 2)2 2 √ 2 π ∝ e−X cos2 Xρ 2 − 4 −iπ/4 iπ/4 P (P ) ∝ |e ϕα (P ) + e ϕ−α (P )|2 √ 2 √ exp(−(P − ρ 2) ) + exp(−(P + ρ 2)2 ) . In the latter equation, we√have used the fact that, for ρ 1, the two Gaussians √ centered at ρ 2 and −ρ 2 have a negligible overlap. 12.3.2. Alice will ﬁnd two peaks, each of which contains roughly half of the events, centered respectively at p0 and −p0 .

118

12 Schr¨ odinger’s Cat

12.3.3. The statistical mixture of Bob leads to the same momentum distribution as that measured by Alice: the N/2 oscillators in the state |α all lead to a mean momentum +p0 , and the N/2 oscillators in the state | − α to −p0 . Up to this point, there is therefore no diﬀerence, and no paradoxical behavior related to the quantum superposition (12.3). 12.3.4. In the X variable, the resolution of the detector satisﬁes δX

1 1 = |α| ρ

Alice therefore has √ a suﬃcient resolution to observe the oscillations of the function cos2 (Xρ 2 − π/4) in the distribution P (X). The shape of the distribution will therefore reproduce the probability law of X drawn on ﬁgure 12.1, i.e. a modulation of period (¯ hπ 2 /(2mα2 ω))1/2 , with a Gaussian envelope. 12.3.5. If Bob performs a position measurement on the N/2 systems in the state |α, he will ﬁnd a Gaussian distribution corresponding to the probability law P (X) ∝ |ψα (X)|2 ∝ exp(−X)2 . He will ﬁnd the same distribution for the N/2 systems in the state | − α. The sum of his results will be a Gaussian distribution, which is quite diﬀerent from the result expected by Alice. The position measurement should, in principle, allow one to discriminate between the quantum superposition and the statistical mixture. 1 h ¯ −26 m. Unfortu12.3.6. The necessary resolution is δx |α| mω ∼ 5 10 nately, it is impossible to attain such a resolution in practice. Section 12.4: The Fragility of a Quantum Superposition 12.4.1. We have E(t) = h ¯ ω(|α0 |2 e−2γt + 1/2): this energy decreases with time. After a time much longer than γ −1 , the oscillator is in its ground state. This dissipation model corresponds to a zero temperature environment. The mean energy acquired by the environment E(0)−E(t) is, for 2γt 1, ∆E(t) 2¯hω|α0 |2 γt. 12.4.2. (a) The probability distribution of the position keeps its Gaussian envelope, but the contrast of the oscillations is reduced by a factor η. (b) The probability distribution of the momentum is given by P(p) =

1 (|ϕα1 (p)|2 + |ϕ−α1 (p)|2 + 2η Re(iϕ∗−α1 (p)ϕα1 (p))) 2

Since the overlap of the two Gaussians ϕα1 (p) and ϕ−α1 (p) is negligible for |α1 | 1, the crossed term, which is proportional to η√ does not contribute hω. signiﬁcantly. One recovers two peaks centered at ±|α1 | 2m¯ The diﬀerence between a quantum superposition and a statistical mixture can be made by position measurements. The quantum superposition leads to a modulation of spatial period (¯ hπ 2 /(2mα2 ω))1/2 with a Gaussian envelope,

12.6 Comments

119

whereas only the Gaussian is observed on a statistical mixture. In order to see this modulation, it must not be too small, say η ≥ 1/10 . 12.4.3. (a) A simple calculation gives β| − β = e−|β|

2

β ∗n (−β)n n

n!

2

2

= e−|β| e−|β| = e−2|β|

2

2

(b) From the previous considerations, we must have e−2|β| ≥ 1/10, i.e. |β| ≤ 1. For times shorter than γ −1 , the energy of the ﬁrst oscillator is E(t) = E(0) − 2γt|α0 |2 ¯hω . The energy of the second oscillator is ¯ ω(|β(t)|2 + 1/2) = h ¯ ω/2 + 2γt|α0 |2 ¯hω . E (t) = h The total energy is conserved; the energy transferred during time t is ∆E(t) = hω = h ¯ ω|β|2 . In order to distinguish between a quantum superposi2γt|α0 |2 ¯ tion and a statistical mixture, we must have ∆E ≤ ¯hω. In other words, if a single energy quantum h ¯ ω is transferred, it becomes problematic to tell the diﬀerence. 12.4.4. With 1/2γ = 1 year = 3 × 107 seconds, the time it takes to reach |β| = 1 is (2γ|α0 |2 )−1 2 × 10−12 seconds!

12.6 Comments Even for a system as well protected from the environment as what we have assumed for the pendulum, the quantum superpositions of macroscopic states are unobservable. After a very short time, all measurements one can make on a system initially prepared in such a state coincide with those made on a statistical mixture. It is therefore not possible, at present, to observe the eﬀects related to the paradoxical character of a macroscopic quantum superposition. However, it is quite possible to observe “mesoscopic” kittens, for systems which have a limited number of degrees of freedom and are well isolated. The ﬁrst attempts concerned SQUIDS (Josephson junctions in superconducting rings), but the results were not conclusive. The idea developed here is oriented towards quantum optics, and has been proposed by Bernard Yurke and David Stoler, Phys. Rev. Lett. 57, p. 13 (1986). The most conclusive results have been obtained at the Ecole Normale Superieure in Paris, on microwave photons (50 GHz) stored in a superconducting cavity (M. Brune, E. Hagley, J. Dreyer, X. Maitre, A. Maali, C. Wunderlich, J.-M. Raimond, and S. Haroche, Phys.

120

12 Schr¨ odinger’s Cat

Rev. Lett. 77, 4887 (1996)). The ﬁeld stored in the cavity is a quasi-perfect harmonic oscillator. The preparation of the kitten (Sect. 2) is accomplished by sending atoms through the cavity. Dissipation (Sect. 4) corresponds to the very weak residual absorption by the walls of the superconducting cavity. One can devise “kittens” made of 5 or 10 photons (i.e. |α|2 = 5 or 10) and one can check precisely the theory, including the decoherence due to dissipation eﬀects.

13 Quantum Cryptography

Cryptography consists in sending a message to a correspondent and in minimizing the risk for this message to be intercepted by an unwanted outsider. The present chapter shows how quantum mechanics can provide a procedure to achieve this goal. We assume here that Alice (A) wants to send Bob (B) some information which may be coded in the binary system, for instance + + − − − + + − ···

(13.1)

We denote the number of bits of this message by n. Alice wants to send this message to Bob only if she has made sure that no “spy” is listening to the communication.

13.1 Preliminaries ˆ = (¯ ˆ where the set Consider a spin 1/2 particle. The spin operator is S h/2)σ σ ˆi , i = x, y, z are the Pauli matrices. We write |σz = +1 and |σz = −1 for h/2. the eigenstates of Sˆz with respective eigenvalues +¯h/2 and −¯ Consider a particle in the state |σz = +1. One measures the component of the spin along an axis u in the (x, z) plane, deﬁned by the unit vector eu = cos θ ez + sin θ ex ,

(13.2)

where ez and ex are the unit vectors along the z and x axes respectively. We recall that the corresponding operator is ¯ ˆ · eu = h (cos θ σ ˆz + sin θ σ ˆx ) . S 2

(13.3)

13.1.1. Show that the possible results of the measurement are +¯h/2 and −¯ h/2.

122

13 Quantum Cryptography

13.1.2. Show that the eigenstates of the observable (13.3) are (up to a multiplicative constant): |σu = +1 = cos φ |σz = +1 + sin φ |σz = −1 |σu = −1 = − sin φ |σz = +1 + cos φ |σz = −1 h/2 and and express φ in terms of θ. Write the probabilities p± u of ﬁnding +¯ −¯ h/2 when measuring the projection of the spin along the u axis. 13.1.3. What are the spin states after measurements that give the results +¯ h/2 and −¯h/2 along u? 13.1.4. Immediately after such a measurement, one measures the z component of the spin. (a) What are the possible results and what are the probabilities of ﬁnding these results in terms of the results found previously along the u axis (observable (13.3)). h/2 as in (b) Show that the probability to recover the same value Sz = +¯ the initial state |σz = +1 is P++ (θ) = (1 + cos2 θ)/2. (c) Assuming now that the initial state is |σz = −1, what is, for the same h/2 sequence of measurements, the probability P−− (θ) to recover Sz = −¯ in the last measurement?

Fig. 13.1. A source emits a pair (a, b) of spin-1/2 particles. Alice measures the component of the spin of a along a direction θa and Bob measures the component of the spin of b along a direction θb

13.2 Correlated Pairs of Spins A source produces a pair (a, b) of spin-1/2 particles (Fig. 13.1), prepared in the state |ψ = φ(r a , r b )|Σ where the spin state of the two particles is 1 |Σ = √ (|σza = +1 ⊗ |σzb = +1 + |σza = −1 ⊗ |σzb = −1) . 2

(13.4)

13.2 Correlated Pairs of Spins

123

Fig. 13.2. A spy, sitting between the source and Bob, measures the component of the b spin along an axis θs

In other words, the spin variables are decoupled from the space variables (r a , r b ). In (13.4), |σua = ± (speciﬁcally u = z) are the eigenstates of the u component of the spin of particle a, and similarly for b. 13.2.1. Show that this state can also be written as: 1 |Σ = √ (|σxa = +1 ⊗ |σxb = +1 + |σxa = −1 ⊗ |σxb = −1) . 2

(13.5)

13.2.2. The pair of particles (a, b) is prepared in the spin state (13.4), (13.5). As the two particles move away from each another, this spin state remains unchanged (unless a measurement is made). (a) Alice ﬁrst measures the spin component of a along an axis ua of angle θa . What are the possible results and the corresponding probabilities in the two cases θa = 0, i.e. the z axis, and θa = π/2, i.e. the x axis? (b) Show that, after Alice’s measurement, the spin state of the two particles depends as follows on the measurement and its result Axis z z x

Result +¯ h/2 +¯ h/2 −¯ h/2

State |σza |σxa |σxa

= +1 ⊗ |σzb = +1 = +1 ⊗ |σxb = +1 = −1 ⊗ |σxb = −1

¿From then on, why can one ignore particle a as far as spin measurements on b are concerned? ˆ = A⊗ ˆ B, ˆ where A ˆ and (We recall that if |ψ = |u⊗|v is a factorized state and C ˆ act respectively on the spaces of |u and |v, then ψ|C|ψ ˆ ˆ ˆ B = u|A|uv| B|v).

13.2.3. After Alice’s measurement, Bob measures the spin of particle b along an axis ub of angle θb . Give the possible results of Bob’s measurement and their probabilities in terms of Alice’s results in the four following conﬁgurations: (a) θa = 0, θb = 0; (b) θa = 0, θb = π/2; (c) θa = π/2, θb = 0;

124

13 Quantum Cryptography 1. Alice and Bob decide along which axes x and z they will make their measurements. 2. Alice, who controls the source S, prepares an ordered sequence of N n pairs of spins in the state (13.4) (n is the number of bits of the message). She sends the b spins to Bob and keeps the a spins. 3. For each spin that they collect, Alice and Bob measure either the x or the z component. Each of them chooses the x or z direction at random with probability p = 1/2. There is no correlation, for a given pair of spins (a, b), between the axis chosen by Alice and the one chosen by Bob. They both register all their results. 4. Bob selects a subset F N of his measurements. He communicates openly to Alice (by cell phone, www, etc.) the axis and the result of the measurement for each event of this subset. In practice F ∼ 0.5. 5. Alice compares, for this subset F N , her axes and her results with those just communicated by Bob. By doing so, she can tell whether or not a spy is present. If a spy is spotted, the procedure stops and a “physical” search for the spy must be undertaken. Otherwise: 6. Alice makes a public announcement that she is convinced not to have been spied upon, and Bob, still openly, communicates his axes of measurements for the remaining spins. However, he does not communicate the corresponding results. 7. . . . Fig. 13.3. The procedure for quantum cryptography

(d) θa = π/2, θb = π/2. In which cases do the measurements on a and b give with certainty the same result? 13.2.4. Consider the situation θa = 0. Suppose that a “spy” sitting between the source and Bob measures the spin of particle b along an axis us of angle θs as sketched in Fig. 13.2. (a) What are, in terms of θs and of Alice’s ﬁndings, the results of the spy’s measurements and their probabilities? (b) After the spy’s measurement, Bob measures the spin of b along the axis deﬁned by θb = 0. What does Bob ﬁnd, and with what probabilities, in terms of the spy’s results? (c) What is the probability P (θs ) that Alice and Bob ﬁnd the same results after the spy’s measurement? (d) What is the expectation value of P (θs ) if the spy chooses θs at random in the interval [0, 2π] with uniform probability? What is this expectation value if the spy chooses only the two values θs = 0 and θs = π/2 each with the same probability p = 1/2?

13.3 The Quantum Cryptography Procedure

125

13.3 The Quantum Cryptography Procedure In order to transmit conﬁdential information, Alice and Bob use the procedure outlined in Fig. 13.3. Comment on this procedure, and answer the following questions. 13.3.1. How can Alice be sure that a spy is present? 13.3.2. What is the probability that an operating spy will escape being detected? Calculate this probability for F N = 200. 13.3.3. Does the spy become more “invisible” if he knows the system of axes (x, z) chosen by Alice and Bob to perform their measurements? 13.3.4. Comment on the two “experiments” whose results are given in Tables 13.1 and 13.2. Show that a spy has certainly listened to communication 2. What is the probability that a spy listened to communication 1, but remained undetected? 13.3.5. Complete the missing item (number 7 in the above procedure), and indicate how Alice can send her message (13.1) to Bob without using any other spin pairs than the N pairs which Bob and her have already analyzed. Using Table 13.3, tell how, in experiment 1, Alice can send to Bob the message (+, −).

Table 13.1. Experiment 1, performed with 12 pairs of spins. Top: set of axes and results obtained by Alice. Bottom: choices of axes and results publicly communicated by Bob A Spin # 1 2 3 4 5 6 7 8 9 10 11 12 A Axis x x z x z z x z z z x x A Result + − + + − − + + + − + − B Spin # 1 2 3 4 5 6 7 8 9 10 11 12 B Axis x x z x x x B Result + − − + + + Table 13.2. Experiment 2, performed with 12 pairs of spins. Top: set of axes and results obtained by Alice. Bottom: choices of axes and results publicly communicated by Bob A Spin # 1 2 3 4 5 6 7 8 9 10 11 12 A Axis x z z z x x z x x z x z A Result + + − + + − + + − − + + B Spin # 1 2 3 4 5 6 7 8 9 10 11 12 B Axis x x x z z z B Result + + − + + −

126

13 Quantum Cryptography

Table 13.3. Choice of axes publicly communicated by Bob in the framework of experiment 1, after Alice has said she is convinced that she is not being spied upon Spin # 2 5 6 8 9 12 Axis x x x z x x

13.4 Solutions Section 13.1: Preliminaries 13.1.1. The spin observable along the u axis is

h cos θ sin θ ˆ · eˆu = ¯ S . 2 sin θ − cos θ ˆ · eˆu , i.e. The possible results of the measurement are the eigenvalues of S ±¯ h/2. 13.1.2. The corresponding eigenvectors are cos(θ/2)|σz = +1 + sin(θ/2)|σz = −1 |σu = +1 = |σu = −1 = − sin(θ/2)|σz = +1 + cos(θ/2)|σz = −1 , therefore φ = θ/2. The probabilities follow directly: 2 2 + 2 − p± u = |σu = ±1|σz = +1| , pu = cos (θ/2) , pu = sin (θ/2) .

13.1.3. The state after a measurement with the result +¯h/2 (or −¯ h/2) is |σu = +1 (or |σu = −1). 13.1.4. (a) If the measurement along u has given +¯h/2, then the probabilities for the second measurement are: h/2) = |σz = ±1|σu = +1|2 p+ z (±¯ with h/2) = cos2 (θ/2) , p+ h/2) = sin2 (θ/2) . p+ z (+¯ z (−¯ If the measurement along u has given −¯ h/2, then h/2) = cos2 (θ/2) , p− h/2) = sin2 (θ/2) . p− z (−¯ z (+¯ h/2 with probabilities: (b) One recovers Sz = +¯ + (i) p+ h/2) = cos4 (θ/2) if the measurement along u has given +¯h/2, u .pz (+¯ − − h/2) = sin4 (θ/2) if the measurement along u has given −¯ h/2. (ii) pu .pz (+¯

13.4 Solutions

127

Altogether, one has P++ = cos4

1 θ θ + sin4 = (1 + cos2 θ) . 2 2 2

(c) The intermediate results are reversed, but the ﬁnal probability is the same 1 P−− = (1 + cos2 θ) . 2 Section 13.2: Correlated Pairs of Spins 13.2.1. The√z and x eigenstates are related by |σx = ±1 = (|σz = +1 ± |σz = −1)/ 2 . If we make the substitution in expression (13.4), we obtain 1 √ (|σza = +1 + |σza = −1) ⊗ (|σzb = +1 + |σzb = −1) 2 2 + (|σza = +1 − |σza = −1) ⊗ (|σzb = +1 − |σzb = −1) , where the crossed terms disappear. More generally, the state under consideration is actually invariant under rotations around the y axis. In an actual experiment, it would be simpler to work with the singlet state √ 1 |0, 0 = √ (|σza = +1 ⊗ |σzb = −1 − |σza = −1 ⊗ |σzb = +1)/ 2 , 2 where Alice and Bob would simply ﬁnd results of opposite signs by measuring along the same axis. 13.2.2. (a) Alice ﬁnds ±¯ h/2 with p = 1/2 in each case. This result is obtained by noticing that the projector on the eigenstate |σza = +1 is Pˆ+a = |σza = +1σza = +1| ⊗ Iˆb and that p(+¯ h/2) = Σ|Pˆ+a |Σ = 1/2, (and similarly for p(−¯ h/2)). (b) This array of results is a consequence of the reduction of the wave packet. If Alice measures along the z axis, we use (13.4); the normalized projections on the eigenstates of Sˆza are |σza = +1 ⊗ |σzb = +1 (Alice’s result: +¯h/2) h/2). A similar formula holds for and |σza = −1 ⊗ |σzb = −1 (Alice’s result: −¯ a measurement along the x axis, because of the invariance property, and its consequence, (13.5). Any measurement on b (a probability, an expectation value) will imply ˆ b where B ˆ b is a projector expectation values of operators of the type Iˆa ⊗ B or a spin operator. Since the states under consideration are factorized, the corresponding expressions for spin measurements on b will be of the type ˆ b (|σ a = +1 ⊗ |σ b = +1) . (σza = +1| ⊗ σzb = +1|)Iˆa ⊗ B z z

128

13 Quantum Cryptography

This reduces to ˆ b |σ b = +1 = σ b = +1|B ˆ b |σ b = +1 σza = +1|σza = +1.σzb = +1|B z z z where the spin state of a is irrelevant. 13.2.3. For the ﬁrst and second conﬁgurations, we can summarize the results as follows: θa

θb

Alice

Bob

Probability

0 0 0 0

0 0 π/2 π/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

+¯h/2 −¯ h/2 ±¯ h/2 ±¯ h/2

p=1 p=1 p± = 1/2 p± = 1/2

The results for θa = π/2, θb = 0 are identical to those of θa = 0, θb = π/2; similarly, the case θa = π/2, θb = π/2 is identical to θa = 0, θb = 0 (one actually recovers the same result for any θa = θb ). In the two cases (a) and (d), where θa = θb , i.e. when they measure along the same axis, Alice and Bob are sure to ﬁnd the same result. 13.2.4. (a) Concerning the ﬁndings of Alice and of the spy, we have: Alice

Spy

Probability

+¯ h/2 +¯ h/2 −¯ h/2 −¯ h/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

cos2 (θs /2) sin2 (θs /2) sin2 (θs /2) cos2 (θs /2)

(b) Concerning the ﬁndings of Bob and of the spy: Spy

Bob

Probability

+¯ h/2 +¯ h/2 −¯ h/2 −¯ h/2

+¯h/2 −¯ h/2 +¯h/2 −¯ h/2

cos2 (θs /2) sin2 (θs /2) sin2 (θs /2) cos2 (θs /2)

(c) The probability that Alice and Bob ﬁnd the same result has actually been calculated in questions 1.4(b,c), we simply have 1 (1 + cos2 θs ) . 2 (d) Amazingly enough, the two expectation values are the same. On one 2π hand, one has 0 P (θs )dθs /(2π) = 3/4. On the other, since P (0) = 1 and P (π/2) = 1/2, on the average p¯ = 3/4 if the values θs = 0 and θs = π/2 are chosen with equal probabilities. P (θS ) =

13.4 Solutions

129

Section 13.3: The Quantum Cryptography Procedure 13.3.1. Necessarily, if θa = θb , the results of Alice and Bob must be the same. If a single measurement done along the same axis θa = θb gives diﬀerent results for Alice and Bob, a spy is certainly operating (at least in an ideal experiment). If θa = θb , on the average half of the results are the same, half have opposite signs. 13.3.2. The only chance for the spy to remain invisible is that Alice and Bob always ﬁnd the same results when they choose the same axis. For each pair of spins, there is a probability 1/2 that they choose the same axis, and there is in this case a probability 1/4 that they do not ﬁnd the same result if a spy is operating (question 2.4(d)). Therefore, for each pair of spins, there is a probability 1/8 that the spy is detected, and a probability 7/8 that the spy remains invisible. This may seem a quite ineﬃcient detection method. However, for a large FN that the spy remains undetected number of events, the probability (7/8) 200 ≈ 2.5 × 10−12 . is very small. For F N = 200 one has (7/8) 13.3.3. Quite surprisingly, as mentioned above, the spy does not gain anything in ﬁnding out which x and z axes Alice and Bob have agreed on in step 1 of the procedure. 13.3.4. Experiment number 2. Measurements 8 and 12, where the axes are the same, give opposite results: rush upon the spy! In experiment number 1, however, measurements 1, 7 and 11 along the x axis do give the same results and are consistent with the assumption that there is no spy around. However, the number N = 3 is quite small in the present case. If a spy is operating, the probability that he remains undetected is ≈ 40%. 13.3.5. Among the (1 − F )N remaining measurements, Alice selects a sequence of events where the axes are the same and which reproduces her message. She communicates openly to Bob the labels of these events, and Bob can (at last!) read the message on his own set of data. In the present case, Alice tells Bob to look at the results # 8 and # 12, where Bob can read (+, −). Comment: This procedure is presently being developed in several industrial research laboratories. In practice, one uses photon pairs with correlated polarizations rather than spin 1/2 particles. See, for instance C. Bennett, G. Brassard, and A. Ekert, Quantum Cryptography, Scientiﬁc American, Vol. 267, p. 26 (October 1992).

14 Direct Observation of Field Quantization

We consider here a two-level atom interacting with a single mode of the electromagnetic ﬁeld. When this mode is treated quantum mechanically, speciﬁc features occur in the atomic dynamics, such as damping and revivals of the Rabi oscillations.

14.1 Quantization of a Mode of the Electromagnetic Field We recall that in classical mechanics, a harmonic oscillator of mass m and frequency ω/2π obeys the equations of motion dx/dt = p/m and dp/dt = Deﬁn−m ω 2 x where x is the position and p the momentum of the oscillator. √ ing the reduced variables X(t) = x(t) mω/¯h and P (t) = p(t)/ ¯hmω, the equations of motion of the oscillator are dP = −ωX , dt

dX = ωP dt

(14.1)

and the total energy U (t) is given by ¯ω 2 h (X (t) + P 2 (t)) . (14.2) 2 14.1.1. Consider a cavity for electromagnetic waves, of volume V . Throughout this chapter, we consider a single mode of the electromagnetic ﬁeld, of the form B(r, t) = uy b(t) cos kz , E(r, t) = ux e(t) sin kz U (t) =

where ux , uy and uz are an orthonormal basis. We recall Maxwell’s equations in vacuum: ∇ · E(r, t) = 0 ∇ · B(r, t) = 0

∂B(r, t) ∂t 1 ∂E(r, t) ∇ ∧ B(r, t) = 2 c ∂t

∇ ∧ E(r, t) = −

132

14 Direct Observation of Field Quantization

and the total energy U (t) of the ﬁeld in the cavity:

0 2 1 2 E (r, t) + U (t) = B (r, t) d3 r with 0 µ0 c2 = 1 . 2 2µ0 V

(14.3)

(a) Express de/dt and db/dt in terms of k, c, e(t), b(t). (b) Express U (t) in terms of V, e(t), b(t), 0 , µ0 . One can take V . sin2 kz d3 r = cos2 kz d3 r = 2 V V (c) Setting ω = ck and introducing the reduced variables χ(t) =

0 V e(t) 2¯ hω

Π(t) =

V b(t) 2µ0 ¯hω

show that the equations for dχ/dt, dΠ/dt and U (t) in terms of χ, Π and ω are formally identical to equations (14.1) and (14.2). 14.1.2. The quantization of the mode of the electromagnetic ﬁeld under consideration is performed in the same way as that of an ordinary harmonic oscillator. One associates to the physical quantities χ and Π, Hermitian operators ˆ which satisfy the commutation relation χ ˆ and Π ˆ =i. [χ, ˆ Π] The Hamiltonian of the ﬁeld in the cavity is ¯ω 2 ˆ2 . ˆC = h χ ˆ +Π H 2 The energy of the ﬁeld is quantized: En = (n + 1/2) h ¯ ω (n is a non-negative ˆ C with eigenvalue En . integer); one denotes by |n the eigenstate of H The quantum states of the ﬁeld in the cavity are linear combinations of the ¯ ω/2, is called the “vacuum”, and the set {|n}. The state |0, of energy E0 = h hω is called the “n photon state”. A “photon” state |n of energy En = E0 + n¯ corresponds to an elementary excitation of the ﬁeld, of energy h ¯ ω. One introduces√the “creation” and “annihilation” operators of a photon as √ ˆ ˆ ˆ − iΠ)/ 2 and a ˆ = (χ ˆ + iΠ)/ 2 respectively. These operators satisfy a ˆ† = (χ the usual relations: √ a ˆ† |n = n + 1|n + 1 √ a ˆ|n = n|n − 1 if n = 0 and a ˆ|0 = 0 . ˆ =a ˆ C in terms of a ˆ† and a ˆ. The observable N ˆ† a ˆ is called the (a) Express H “number of photons”.

14.2 The Coupling of the Field with an Atom

133

The observables corresponding to the electric and magnetic ﬁelds at a point r are deﬁned as:

hω ¯ ˆ a ˆ+a ˆ† sin kz E(r) = ux 0 V

µ0 ¯ hω † ˆ a ˆ −a ˆ cos kz . B(r) = iuy V The interpretation of the theory in terms of states and observables is the same as in ordinary quantum mechanics. ˆ C |n in an (b) Calculate the expectation values E(r), B(r), and n|H n-photon state. 14.1.3. The following superposition: |α = e

−|α|2 /2

∞ αn √ |n , n! n=0

(14.4)

where α is any complex number, is called a “quasi-classical” state of the ﬁeld. (a) Show that |α is a normalized eigenvector of the annihilation operator a ˆ and give the corresponding eigenvalue. Calculate the expectation value n of the number of photons in that state. (b) Show that if, at time t = 0, the state of the ﬁeld is |ψ(0) = |α, then, at time t, |ψ(t) = e−iωt/2 |(αe−iωt ). (c) Calculate the expectation values E(r)t and B(r)t at time t in a quasi-classical state for which α is real. (d) Check that E(r)t and B(r)t satisfy Maxwell’s equations. (e) Calculate the energy of a classical ﬁeld such that E cl (r, t) = E(r)t ˆ and B cl (r, t) = B(r) t . Compare the result with the expectation value ˆ of HC in the same quasi-classical state. (f) Why do these results justify the name “quasi-classical” state for |α if |α| 1?

14.2 The Coupling of the Field with an Atom Consider an atom at point r 0 in the cavity. The motion of the center of mass of the atom in space is treated classically. Hereafter we restrict ourselves to the two-dimensional subspace of internal atomic states generated by the ground state |f and an excited state |e. The origin of atomic energies is chosen in such a way that the energies of |f and |e are respectively −¯ hωA /2 and +¯ hωA /2 (ωA > 0). In the basis {|f , |e}, one can introduce the operators:

1 0 0 0 0 1 σ ˆz = σ ˆ+ = σ ˆ− = , 0 −1 1 0 0 0

134

14 Direct Observation of Field Quantization

that is to say σ ˆ+ |f = |e and σ ˆ− |e = |f , and the atomic Hamiltonian can ˆ A = − h¯ ωA σ ˆ . be written as: H z 2 The set of orthonormal states {|f, n , |e, n, n ≥ 0} where |f, n ≡ |f ⊗|n and |e, n ≡ |e⊗|n forms a basis of the Hilbert space of the {atom+photons} states. ˆA + H ˆ C , and give the correˆ0 = H 14.2.1. Check that it is an eigenbasis of H sponding eigenvalues. 14.2.2. In the remaining parts of the problem we assume that the frequency of the cavity is exactly tuned to the Bohr frequency of the atom, i.e. ω = ωA . ˆ 0 . Show that, Draw schematically the positions of the ﬁrst 5 energy levels of H ˆ 0 are grouped in degenerate except for the ground state, the eigenstates of H pairs. 14.2.3. The Hamiltonian of the electric dipole coupling between the atom and the ﬁeld can be written as:

ˆ =γ a W ˆσ ˆ+ + a ˆ† σ ˆ− , where γ = −d ¯hω/0 V sin kz0 , and where the electric dipole moment d is determined experimentally. ˆ on the states |f, n and |e, n. (a) Write the action of W ˆ† σ ˆ− correspond? (b) To which physical processes do a ˆσ ˆ+ and a ˆ = H ˆ0 + W ˆ and the corresponding 14.2.4. Determine the eigenstates of H energies. Show that the problem reduces to the diagonalization of a set of 2 × 2 matrices. One hereafter sets: 1 |φ± n = √ (|f, n + 1 ± |e, n) 2 hΩ0 ¯ hω ¯ = γ = −d sin kz0 2 0 V

√ Ω n = Ω0 n + 1 .

± The energies corresponding to the eigenstates |φ± n are denoted En .

14.3 Interaction of the Atom with an “Empty” Cavity In the following, one assumes that the atom crosses the cavity along a line where sin kz0 = 1. An atom in the excited state |e is sent into the cavity prepared in the vacuum state |0. At time t = 0 , when the atom enters the cavity, the state of the system is |e, n = 0. 14.3.1. What is the state of the system at a later time t?

14.4 Interaction of an Atom

with a Quasi-Classical State

135

14.3.2. What is the probability Pf (T ) to ﬁnd the atom in the state f at time T when the atom leaves the cavity? Show that Pf (T ) is a periodic function of T (T is varied by changing the velocity of the atom). 14.3.3. The experiment has been performed on rubidium atoms for a couple of states (f, e) such that d = 1.1 × 10−26 C.m and ω/2π = 5.0 × 1010 Hz. The volume of the cavity is 1.87 × 10−6 m3 (we recall that 0 = 1/(36π109 ) S.I.). The curve ∞ Pf (T ), together with the real part of its Fourier transform J(ν) = 0 cos (2πνT ) Pf (T ) dT , are shown in Fig. 14.1. One observes a damped oscillation, the damping being due to imperfections of the experimental setup. How do theory and experiment compare? (We recall that the Fourier transform of a damped sinusoid in time exhibits a peak at the frequency of this sinusoid, whose width is proportional to the inverse of the characteristic damping time.)

1,0

Pf (t)

0,04

(a)

0,8

J(ν)

(b)

0,03

0,6

0,02

0,4 0,01

0,2 0,0

t (µs) 0

20

40

60

80

100

ν (kHz) 0,00

0

20

40

60

80 100 120 140

Fig. 14.1. (a) Probability Pf (T ) of detecting the atom in the ground state after it crosses a cavity containing zero photons; (b) Fourier transform of this probability, as deﬁned in the text

14.4 Interaction of an Atom with a Quasi-Classical State The atom, initially in the state |e, is now sent into a cavity where a quasiclassical state |α of the ﬁeld has been prepared. At time t = 0 the atom enters the cavity and the state of the system is |e ⊗ |α. 14.4.1. Calculate the probability Pf (T, n) to ﬁnd, at time T , the atom in the state |f and the ﬁeld in the state |n + 1, for n ≥ 0. What is the probability to ﬁnd the atom in the state |f and the ﬁeld in the state |0? 14.4.2. Write the probability Pf (T ) to ﬁnd the atom in the state |f , independently of the state of the ﬁeld, as an inﬁnite sum of oscillating functions.

136

14 Direct Observation of Field Quantization

14.4.3. On Fig. 14.2 are plotted an experimental measurement of Pf (T ) and the real part of its Fourier transform J(ν). The cavity used for this measurement is the same as in Fig. 14.1, but the ﬁeld has been prepared in a quasi-classical state before the atom is sent in. (a) Determine the three frequencies ν0 , ν1 , ν2 which contribute most strongly to Pf (T ). (b) Do the ratios ν1 /ν0 and ν2 /ν0 have the expected values? (c) From the values J(ν0 ) and J(ν1 ), determine an approximate value for the mean number of photons |α|2 in the cavity.

Fig. 14.2. (a) Probability Pf (T ) of measuring the atom in the ground state after the atom has passed through a cavity containing a quasi-classical state of the electromagnetic ﬁeld; (b) Fourier transform of this probability

14.5 Large Numbers of Photons: Damping and Revivals Consider a quasi-classical state |α of the ﬁeld corresponding to a large mean number of photons: |α|2 n0 1, where n0 is an integer. In this case, the probability π(n) to ﬁnd n photons can be cast, in good approximation, in the form:

2n 2 |α| 1 (n − n0 )2 √ exp − π(n) = e−|α| . n! 2 n0 2πn0 This Gaussian limit of the Poisson distribution can be obtained by using the √ Stirling formula n! ∼ nn e−n 2πn and expanding ln π(n) in the vicinity of n = n0 . 14.5.1. Show that this probability takes signiﬁcant values only if n lies in a neighborhood δn of n0 . Give the relative value δn/n0 . 14.5.2. For such a quasi-classical state, one tries to evaluate the probability Pf (T ) of detecting the atom in the state f after its interaction with the ﬁeld. In order to do this,

14.6 Solutions

137

• one linearizes the dependence of Ωn on n in the vicinity of n0 : n − n0 Ωn Ωn0 + Ω0 √ , 2 n0 + 1

(14.5)

• one replaces the discrete summation in Pf (T ) by an integral. (a) Show that, under these approximations, Pf (T ) is an oscillating function of T for short times, but that this oscillation is damped away after a characteristic time TD . Give the value of TD . We recall that ∞ 2 2 2 2 1 √ e−(x−x0 ) /2σ cos(αx) dx = e−α σ /2 cos(αx0 ). −∞ σ 2π (b) Does this damping time depend on the mean value of the number of photons n0 ? (c) Give a qualitative explanation for this damping. 14.5.3. If one keeps the expression of Pf (T ) as a discrete sum, an exact numerical calculation shows that one expects a revival of the oscillations of Pf (T ) for certain times TR large compared to TD , as shown in Fig. 14.3. This phenomenon is called quantum revival and it is currently being studied experimentally. Keeping the discrete sum but using the approximation (14.5), can you explain the revival qualitatively? How does the time of the ﬁrst revival depend on n0 ?

Fig. 14.3. Exact theoretical calculation of Pf (T ) for n 25 photons

14.6 Solutions Section 14.1: Quantization of a Mode of the Electromagnetic Field 14.1.1. (a) The pair of Maxwell equations ∇ · E = 0 and ∇ · B = 0 are satisﬁed whatever the values of the functions e(t) and b(t). The equations

138

14 Direct Observation of Field Quantization

∇ ∧ E = −(∂B/∂t) and c2 ∇ ∧ B = −(∂E/∂t) require that: db = −ke(t) . dt

de = c2 kb(t) dt

(b) The electromagnetic energy can be written as:

0 2 1 2 2 2 e (t) sin kz + b (t) cos kz d3 r U (t) = 2 2µ0 V 0 V 2 V 2 = e (t) + b (t) . 4 2µ0 (c) Under the change of functions suggested in the text, we obtain: !

hω 2 ¯ χ˙ = ω Π U (t) = χ (t) + Π 2 (t) . ˙ Π = −ω χ 2 These two equations are formally identical to the equations of motion of a particle in a harmonic oscillator potential. ˆ = i, we deduce that: 14.1.2. (a) From [χ, ˆ Π] 1 ˆ χ ˆ =1. [χ ˆ + iΠ, ˆ − iΠ] 2 √ √ ˆ = i(ˆ In addition, χ ˆ = (ˆ a+a ˆ† )/ 2 and Π a† − a ˆ)/ 2, i.e.:

hω † 1 ˆC = ¯ H a ˆa ˆ +a ˆ† a ˆ =h ¯ω a ˆ† a ˆ+ , 2 2 ˆ+1 . ˆC = h ¯ω N or H 2 [ˆ a, a ˆ† ] =

(b) For an n photon state, we ﬁnd n|ˆ a|n = n|ˆ a† |n = 0, which results in E(r) = 0

B(r) = 0 .

ˆ C with eigenvalue (n + 1/2)¯ The state |n is an eigenstate of H hω, i.e.

1 HC = n + hω . ¯ 2 14.1.3. (a) The action of a ˆ on |α gives a ˆ|α = e−|α|

2

/2

= αe−|α|

2

∞ αn √ √ n|n − 1 n! n=1

/2

∞ n=1

αn−1 (n − 1)!

|n − 1 = α|α .

14.6 Solutions

139

The vector |α is normalized: α|α = e−|α|

2

∞ (α∗ )n αn =1. n! n=0

The expectation value of the number of photons in that state is: ˆ |α = α|ˆ n = α|N a† a ˆ|α = ||ˆ a|α||2 = |α|2 . (b) The time evolution of |ψ(t) is given by ∞ αn √ e−iω(n+1/2)t |n n! n=0

n ∞ αe−iωt 2 √ |n = e−iωt/2 e−|α| /2 n! n=0

|ψ(t) = e−|α|

2

/2

= e−iωt/2 |(αe−iωt ) . (c) The expectation values of the electric and magnetic ﬁelds are ¯hω ux E(r)t = 2α cos ωt sin kz V 0 ¯hωµ0 uy . B(r)t = −2α sin ωt cos kz V (d) These ﬁelds are of the same type as the classical ﬁelds considered at the beginning of the problem, with hω ¯ ¯hωµ0 cos ωt b(t) = −2α sin ωt . e(t) = 2α 0 V V Given the relation 0 µ0 c2 = 1, we verify that e(t) ˙ = c2 kb(t) and b˙ = −ke(t). Therefore the expectation values of the ﬁeld operators satisfy Maxwell’s equations. (e) The energy of the classical ﬁeld can be calculated using the result of ques¯ ωα2 . This “classical” tion 1.1b. Since cos2 ωt + sin2 ωt = 1, we ﬁnd U (t) = h ˆ C is: energy is therefore time-independent. The expectation value of H hω(N + 1/2) = h ¯ ω(α2 + 1/2) . HC = ¯ It is also time independent (Ehrenfest’s theorem). (f ) For |α| much larger than 1, the ratio U (t)/HC is close to 1. More generally, the expectation value of a physical quantity as calculated for a quantum ﬁeld in the state |α, will be close to the value calculated for a classical ﬁeld such that E cl (r, t) = E(r)t and B cl (r, t) = B(r)t .

140

14 Direct Observation of Field Quantization

Fig. 14.4. (a) Positions of the ﬁve ﬁrst energy levels of H0 . (b) Positions of the ﬁve ˆ ˆ =H ˆ0 + W ﬁrst energy levels of H

Section 14.2: The Coupling of the Field with an Atom 14.2.1. One checks that

1 hωA ¯ + n+ ¯hω |f, n , 2 2

hωA 1 ˆ 0 |e, n = ¯ + n+ H ¯hω |e, n . 2 2

ˆ 0 |f, n = H

−

14.2.2. For a cavity which resonates at the atom’s frequency, i.e. if ω = ωA , ˆ0 the couple of states |f, n + 1, |e, n are degenerate. The ﬁrst ﬁve levels of H are shown in Fig. 14.4a. Only the ground state |f, 0 of the atom+ﬁeld system is non-degenerate. ˆ on the basis vectors of H0 is given by: 14.2.3. (a) The action of W √ ˆ |f, n = n γ |e, n − 1 if n ≥ 1 W = 0 if n = 0 √ ˆ W |e, n = n + 1 γ |f, n + 1 . The coupling under consideration corresponds to an electric dipole interaction ˆ · E(r), ˆ ˆ is the observable electric dipole moment of of the form −D where D the atom. ˆ couples the two states of each degenerate pair. The term a (b) W ˆσ ˆ+ corresponds to the absorption of a photon by the atom, which undergoes a transiˆ− corresponds tion from the ground state to the excited state. The term a ˆ† σ to the emission of a photon by the atom, which undergoes a transition from the excited state to the ground state.

14.6 Solutions

141

ˆ is block-diagonal in the eigenbasis of 14.2.4. The operator W ˆ H0 {|f, n, |e, n} . Therefore: ˆ with the eigenvalue 0. ˆ0 + W • The state |f, 0 is an eigenstate of H ˆ • In each eigen-subspace of H0 generated by {|f, n + 1, |e, n} with n ≥ 0, one must diagonalize the 2 × 2 matrix:

(n + 1)¯ hω ¯ hΩn /2 hω hΩn /2 (n + 1)¯ ¯ whose eigenvectors and corresponding eigenvalues are (n ≥ 0): ¯hΩn 2 ¯hΩn . = (n + 1)¯ hω − 2

|φ+ n

corresponding to

En+ = (n + 1)¯ hω +

|φ− n

corresponding to

En−

ˆ0 + W ˆ are shown in Fig. 14.4b. The ﬁrst ﬁve energy levels of H Section 14.3: Interaction of the Atom and an “Empty” Cavity ˆ 14.3.1. We expand the initial state on on the eigenbasis of H:

1 − |ψ(0) = |e, 0 = √ |φ+ 0 − |φ0 . 2 The time evolution of the state vector is therefore given by: + 1 −iE0− t/¯ h − |φ0 |ψ(t) = √ e−iE0 t/¯h |φ+ 0−e 2 −iωt e iΩ0 t/2 − |φ0 . = √ e−iΩ0 t/2 |φ+ 0−e 2 14.3.2. In general, the probability of detecting the atom in the state f , independently of the ﬁeld state, is given by: Pf (T ) =

∞

|f, n|ψ(T )|2 .

n=0

In the particular case of an initially empty cavity,

√only the term n = 1 con− tributes to the sum. Using |f, 1 = |φ+ 0 + |φ0 / 2, we ﬁnd 1 Ω0 T = (1 − cos Ω0 T ) . 2 2 It is indeed a periodic function of T , with angular frequency Ω0 . Pf (T ) = sin2

14.3.3. Experimentally, one measures an oscillation of frequency ν0 = 47 kHz. This result corresponds to the expected value: hω 1 2d ¯ . ν0 = 2π ¯ h 0 V

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14 Direct Observation of Field Quantization

Section 14.4: Interaction of an Atom with a Quasi-Classical State ˆ: ˆ0 + W 14.4.1. Again, we expand the initial state on the eigenbasis of H |ψ(0) = |e ⊗ |α = e−|α| = e−|α|

2

/2

2

/2

∞ αn √ |e, n n! n=0

∞

αn 1 − √ √ |φ+ n − |φn . n! 2 n=0

At time t the state vector is ∞ 2 + − αn 1 −iEn t/¯ h − √ √ e−iEn t/¯h |φ+ |ψ(t) = e−|α| /2 − e |φ . n n n! 2 n=0 We therefore observe that: • the probability to ﬁnd the atom in the state |f and the ﬁeld in the state |0 vanishes for all values of T , • the probability Pf (T, n) can be√obtained from the scalar product of |ψ(t) − and |f, n + 1 = (|φ+ n + |φn ) / 2: 2 2n 2 |α| − 1 −iEn+ t/¯h Pf (T, n) = e−|α| − e−iEn t/¯h e 4 n! 2n 2n 2 |α| 2 |α| 1 Ωn T sin2 = e−|α| (1 − cos Ωn T ) . = e−|α| n! 2 2 n! 14.4.2. The probability Pf (T ) is simply the sum of all probabilities Pf (T, n): Pf (T ) =

∞ n=0

Pf (T, n) =

2 ∞ 1 e−|α| |α|2n − cos Ωn T . 2 2 n=0 n!

14.4.3. (a) The three most prominent peaks of J(ν) occur at the frequencies ν0 = 47 kHz (already found for an empty cavity), ν1 = 65 kHz and ν2 = 81 kHz. (b) The ratios of the are very close to the theoretical √ measured frequencies √ predictions: ν1 /ν0 = 2 and ν2 /ν0 = 3. (c) The ratio J(ν1 )/J(ν0 ) is of the order of 0.9. Assuming the peaks have the same widths, and that these widths are small compared to the splitting ν1 − ν0 , this ratio correponds to the average number of photons |α|2 in the cavity. Actually, the peaks overlap, which makes this determination somewhat inaccurate. If one performs a more sophisticated analysis, taking into account the widths of the peaks, one obtains |α|2 = 0.85 ± 0.04 (see the reference at end of this chapter). Comment: One can also determine |α|2 from the ratio J(ν2 )/J(ν1 ) which should be equal to |α|2 /2. However, the inaccuracy due to the overlap of the peaks is greater than for J(ν1 )/J(ν0 ), owing to the smallness of J(ν2 ).

14.6 Solutions

143

Section 14.5: Large Numbers of Photons: Damping and Revivals 14.5.1. The probability π(n) takes signiﬁcant values only if (n − n0 )2 /(2n0 ) is not much larger than 1, i.e. for integer values of n in a neighborhood of n0 √ of relative extension of the order of 1/ n0 . For n0 1, the distribution π(n) is therefore peaked around n0 . 14.5.2. (a) Consider the result of question 4.2, where we replace Ωn by its approximation (14.5):

∞ 1 1 n − n0 Pf (T ) = − π(n) cos Ωn0 + Ω0 √ T (14.6) 2 2 n=0 2 n0 + 1 We now replace the discrete sum by an integral: 2 1 1 ∞ e−u /(2n0 ) √ Pf (T ) = − 2 2 −∞ 2πn0

u T du . · cos Ωn0 + Ω0 √ 2 n0 + 1 We have extended the lower integration bound from −n0 down to −∞, using √ the fact that the width of the gaussian is n0 n0 . We now develop the expression to be integrated upon:

Ω uT u √0 T = cos (Ωn0 T ) cos cos Ωn0 +Ω0 √ 2 n0 +1 2 n0 + 1

Ω uT √0 − sin (Ωn0 T ) sin . 2 n0 + 1 The sine term does not contribute to the integral (odd function) and we ﬁnd:

1 1 Ω02 T 2 n0 Pf (T ) = − cos (Ωn0 T ) exp − . 2 2 8(n0 + 1) For n0 1, the argument of the exponential simpliﬁes, and we obtain:

1 1 T2 Pf (T ) = − cos (Ωn0 T ) exp − 2 2 2 TD √ with TD = 2 2/Ω0 . (b) In this approximation, the oscillations are damped out in a time TD which is independent of the number of photons n0 . For a given atomic transition (for ﬁxed values of d and ω), this time TD increases like the square root of the volume of the cavity. In the limit of an inﬁnite cavity, i.e. an atom in empty space, this damping time becomes inﬁnite: we recover the usual Rabi oscillation. For a cavity of ﬁnite size, the number of visible oscillations of √ Pf (T ) is roughly νn0 TD ∼ n0 .

144

14 Direct Observation of Field Quantization

(c) The function Pf (T ) is made up of a large number of oscillating functions with similar frequencies. Initially, these diﬀerent functions are in phase, and their sum Pf (T ) exhibits marked oscillations. After a time TD , the various oscillations are no longer in phase with one another and the resulting oscillation of Pf (T ) is damped. One can ﬁnd the damping time by simply estimating the time for which the two frequencies at half width on either side of the maximum of π(n) are out of phase by π: √ √ 1 and n0 ± n0 n0 ± Ωn0 +√n0 TD ∼ Ωn0 −√n0 TD + π 2 Ω0 T D ∼ π .

⇒

14.5.3. Within the approximation (14.5) suggested in the text, equation (14.6) above corresponds to a periodic evolution of period TR =

4π √ n0 + 1 . Ω0

Indeed

Ωn0

n − n0 + Ω0 √ 2 n0 + 1

TR = 4π (n0 + 1) + 2π(n − n0 ) .

We therefore expect that all the oscillating functions which contribute to Pf (T ) will reset in phase at times TR , 2 TR ,. . . The time of the ﬁrst revival, measured in Fig. 14.3, is Ω0 T 64, in excellent agreement with this predic√ tion. Notice that TR ∼ 4 n0 TD , which means that the revival time is always large compared to the damping time. Actually, one can see from the result of Fig. 14.3 that the functions are only partly in phase. This comes from the fact that the numerical calculation has been done with the exact expression of Ωn . In this case, the diﬀerence between two consecutive frequencies Ωn+1 − Ωn is not exactly a constant, contrary to what happens in approximation (14.5); the function Pf (T ) is not really periodic. After a few revivals, one obtains a complicated behavior of Pf (T ), which can be analysed with the techniques developed for the study of chaos.

14.7 Comments The damping phenomenon which we have obtained above is “classical”: one would obtain it within a classical description of the interaction of the ﬁeld and the atom, by considering a ﬁeld whose intensity is not well deﬁned (this would be the analog of a distribution π(n) of the number of photons). On the other hand, the revival comes from the fact that the set of frequencies Ωn is discrete. It is a direct consequence of the quantization of the electromagnetic

14.7 Comments

145

√ √ ﬁeld, in the same way as the occurrence of frequencies ν0 2, ν0 3,. . . in the evolution of Pf (T ) (Sect. 4). The experiments described in this chapter have been performed in Paris, at the Laboratoire Kastler Brossel. The pair of levels (f, e) correspond to very excited levels of rubidium, which explains the large value of the electric dipole moment d. The ﬁeld is conﬁned in a superconducting niobium cavity (Qfactor of ∼ 108 ), cooled down to 0.8 K in order to avoid perturbations to the experiment due to the thermal black body radiation (M. Brune, F. SchmidtKaler, A. Maali, J. Dreyer, E. Hagley, J.-M. Raimond, and S. Haroche, Phys. Rev. Lett. 76, 1800 (1996)).

15 Ideal Quantum Measurement

In 1940, John von Neumann proposed a deﬁnition for an optimal, or “ideal” measurement of a quantum physical quantity. In this chapter, we study a practical example of such a procedure. Our ambition is to measure the excitation number of a harmonic oscillator S by coupling it to another oscillator D whose phase is measured. We recall that, for k integer: N

e

2iπkn N +1

=N +1

for k = p(N + 1), p integer ;

=0

otherwise.

n=0

15.1 Preliminaries: a von Neumann Detector We want to measure a physical quantity A on a quantum system S. We use a detector D devised for such a measurement. There are two stages in the measurement process. First, we let S and D interact. Then, after S and D get separated and do not interact anymore, we read a result on the detector D. We assume that D possesses an orthonormal set of states {|Di } with Di |Dj = δi,j . These states correspond for instance to the set of values which can be read on a digital display. Let |ψ be the state of the system S under consideration, and |D the state of the detector D. Before the measurement, the state of the global system S + D is |Ψi = |ψ ⊗ |D . Let ai and |φi be the eigenvalues and corresponding eigenstates of the ˆ The state |ψ of the system S can be expanded as observable A. αi |φi . (15.1) |ψ = i

148

15 Ideal Quantum Measurement

15.1.1. Using the axioms of quantum mechanics, what are the probabilities p(ai ) to ﬁnd the values ai in a measurement of the quantity A on this state? 15.1.2. After the interaction of S and D, the state of the global system is in general of the form |Ψf = γij |φi ⊗ |Dj (15.2) i,j

We now observe the state of the detector. What is the probability to ﬁnd the detector in the state |Dj ? 15.1.3. After this measurement, what is the state of the global system S +D? 15.1.4. A detector is called ideal if the choice of |D0 and of the coupling S– D leads to coeﬃcients γij which, for any state |ψ of S, verify: |γij | = δi,j |αj |. Justify this designation.

15.2 Phase States of the Harmonic Oscillator ˆ the We consider a harmonic oscillator of angular frequency ω. We note N ˆ ˆ “number” operator, i.e. the Hamiltonian is H = (N + 1/2))¯ hω with eigenhω, N integer ≥ 0. states |N and eigenvalues EN = (N + 12 )¯ Let s be a positive integer. The so-called “phase states” are the family of states deﬁned at each time t by: N =s 1 √ e−iN (ωt+θm ) |N |θm = s + 1 N =0

(15.3)

where θm can take any of the 2s + 1 values θm =

2πm (m = 0, 1, . . . , s) . s+1

(15.4)

15.2.1. Show that the states |θm are orthonormal. 15.2.2. We consider the subspace of states of the harmonic oscillator such that the number of quanta N is bounded from above by some value s. The sets {|N , N = 0, 1, . . . , s} and {|θm , m = 0, 1, . . . , s} are two bases in this subspace. Express the vectors |N in the basis of the phase states. 15.2.3. What is the probability to ﬁnd N quanta in a phase state |θm ? 15.2.4. Calculate the expectation value of the position x ˆ in a phase state, and ﬁnd a justiﬁcation for the name “phase state”. We recall the relation √ √ − 1), where x is the characteristic x ˆ|N = x0 ( N + 1|N + 1 + N |N 0 √ s length of the problem. We set Cs = N =0 N .

15.4 An “Ideal” Measurement

149

15.3 The Interaction between the System and the Detector We want to perform an “ideal” measurement of the number of excitation quanta of a harmonic oscillator. In order to do so, we couple this oscillator S with another oscillator D, which is our detector. Both oscillators have the ˆ S = (ˆ n + 12 )¯ hω are noted same angular frequency ω. The eigenstates of H 1 ˆ ˆ |n, n = 0, 1, . . . , s, those of HD = (N + 2 )¯ hω are noted |N , N = 0, 1...s ˆ are the number operators of S and D. where n ˆ and N We assume that both numbers of quanta n and N are bounded from above by s. The coupling between S and D has the form: ˆ Vˆ = h ¯g n ˆN

(15.5)

This Hamiltonian is realistic. If the two oscillators are two modes of the electromagnetic ﬁeld, it originates from the crossed Kerr eﬀect .

15.3.1. What are the eigenstates and eigenvalues of the total Hamiltonian ˆ =H ˆS + H ˆ D + Vˆ H

?

15.3.2. We assume that the initial state of the global system S + D is factorized as: |Ψ (0) = |ψS ⊗|ψD , with : |ψS = an |n , |ψD = bN |N (15.6) n

N

where we assume that |ψS and |ψD are normalized. We perform a measurement of n ˆ in the state |Ψ (0). What results can one ﬁnd, with what probabilˆ. ities? Answer the same question for a measurement of N 15.3.3. During the time interval [0, t], we couple the two oscillators. The coupling is switched oﬀ at time t. What is the state |Ψ (t) of the system? Is it also a priori factorizable? 15.3.4. Is the probability law for the couple of random variables {n, N } affected by the interaction? Why?

15.4 An “Ideal” Measurement Initially, at time t = 0, the oscillator S is in a state |ψS = oscillator D is prepared in the state |ψD = √

s 1 |N . s + 1 N =0

s n=0

an |n. The

(15.7)

150

15 Ideal Quantum Measurement

15.4.1. We switch on the interaction Vˆ during the time interval [0, t]. Express the state |Ψ (t) in terms of the phase states {|θk } of the oscillator D. 15.4.2. We assume the interaction time is t = t0 ≡ 2π/ [g(s + 1)]. Write the state |Ψ (t0 ) of the system. 15.4.3. What is the probability to ﬁnd the value θk in a measurement of the phase of the “detector” oscillator D? 15.4.4. After this measurement has been performed, what is the state of the oscillator S? Describe qualitatively what will happen if one were to choose an interaction time t = t0 . 15.4.5. Comment on the result. In your opinion, why did J. von Neumann consider this as an “ideal” quantum-measurement process?

15.5 Solutions Section 15.1: Preliminaries; a Von Neumann Detector 15.1.1. Since the state of the system is |ψ = i αi |φi , the probability to ﬁnd the value aj in a measurement of A is p(aj ) = |αj |2 . 15.1.2. The state of the global system is γij |φi ⊗ |Dj . |Ψ1 = i,j

The probability pj to ﬁnd the detector in the state |Dj is the sum of the probabilities |γij |2 : |γij |2 , pj = i

since the states |φi are orthogonal. 15.1.3. After this measurement, the state of the global system S + D is, after the principle of wave packet reduction, 1 γij |φi ⊗ |Dj . |Ψ = √ pj i 15.1.4. For an ideal detector, the probability that the detector is in the state |Dj is pj = |αj |2 = p(aj ) and the state of the set system + detector, once we know the state of the detector, is |φj ⊗ |Dj . This is the expected result, given the wave packet reduction principle.

15.5 Solutions

151

Section 15.2: Phase states of the harmonic oscillator 15.2.1. Given the deﬁnition of the phase states, one has: θm |θn = = =

s s 1 iN (ωt+θm ) −iN (ωt+θn ) e e N |N s+1

1 s+1 1 s+1

N =0 N =0 s iN (θm −θn )

e

N =0 s

e2iπN (m−n) / (s+1) = δm,n ,

N =0

where the last equality stands because −s ≤ m − n ≤ s. 15.2.2. The scalar product of a state |N with a phase state is θm |N = (N |θm )∗ = √

1 eiN (ωt+θm ) , s+1

hence the expansion: |N =

s

s 1 θm |N |θm = √ eiN (ωt+θm ) |θm . s + 1 m=0 m=0

15.2.3. Given the deﬁnition of a phase state, the probability to ﬁnd N quanta in a state |θm is 1 p(N, θm ) = |N |θm |2 = . s+1 15.2.4. One obtains θm |ˆ x|θm = 2x0

Cs cos (ωt + θm ) . s+1

The phases of the expectation values of x in two phase states |θm and |θn diﬀer by an integer multiple 2(m − n)π/(s + 1) of the elementary phase 2π/(s + 1). Section 15.3: The Interaction between the System and the Detector 15.3.1. The factorized states |n ⊗ |N are eigenstates of the total Hamiltonian ˆ =H ˆS + H ˆ D + Vˆ = (ˆ ˆ + 1)¯ ˆ , H n+N hω + h ¯g n ˆ⊗N with eigenvalues En,N = (n + N + 1)¯ hω + h ¯ g nN . 15.3.2. The results of measurements and the corresponding probabilities are n = 0, 1, . . . , s, p(n) = |an |2 and N = 0, 1, . . . , s, p(N ) = |bN |2 .

152

15 Ideal Quantum Measurement

15.3.3. The state of the system at time t is an bN e−i[(n+N +1)ω+gnN ]t |n ⊗ |N . |Ψ (t) = n

N

In general, it is not factorized. 15.3.4. The probability law for the couple of random variables {n, N } is still p(n, N ) = |an |2 |bN |2 . It is not modiﬁed by the interaction since Vˆ commutes ˆ . The quantities n and N are constants of the motion. with n ˆ and N Section 15.4: An “Ideal” Measurement √ 15.4.1. One has bN = 1/ s + 1, hence 1 an e−i[(n+N +1)ω+gnN )]t |n ⊗ |N . |Ψ (t) = √ s+1 n N Inserting the expansion of the states |N in terms of the phase states, one obtains ei(θm −gnt)N e−i(n+1)ωt an |n ⊗ |θm . |Ψ (t) = s + 1 n m N

15.4.2. If the interaction time is t0 = 2π / [g(s + 1)], this expression reduces to s e−i(n+1)ωt0 an |n ⊗ |θn . (15.8) |Ψ (t0 ) = n=0

15.4.3. The probability to ﬁnd the result θn by measuring the phase of the detecting oscillator D on this state is p(θn ) = |an |2 . 15.4.4. After this measurement, the state of the oscillator S is simply |n (up to an arbitrary phase factor). In the state (15.8), the two systems are perfectly correlated. To a phase state of D there corresponds only one state of number of quanta of S. If one were to choose a time interval diﬀerent from t0 , this correlation would not be perfect. After a measurement of the phase of D, the state of S would be a superposition of states with diﬀerent numbers of quanta. 15.4.5. We see that this procedure, which supposes a well deﬁned interaction time interval between the system and the detector, gives the value of the probability p(n) = |an |2 that S is in a state with n quanta. In addition, after one has read the result θn on the detector, one is sure that S is in the state |n, without having to further interact with it (reduction of the wave packet). In this sense, this procedure does follow exactly the axioms of quantum mechanics on measurement. It is therefore an “ideal” measurement of a quantum physical quantity.

15.6 Comments

153

15.6 Comments One can extend formally this result to other systems than harmonic oscillators. In practice, the case studied here is a simpliﬁcation of the concrete case where the oscillators S and D are modes of the electromagnetic ﬁeld. The Hamiltonian which is eﬀectively encountered in a optically non-linear crystal, comes from the phenomenon called the crossed Kerr eﬀect . In an interferometer, where D is a laser beam split in two parts by a semi-transparent mirror, one can let the signal oscillator S interact with one of the beams. The measurement consists in an interferometric measurement when the two beams of D recombine. This type of experiment has been carried out intensively in recent years. It is also called a “non-destructive” quantum measurement (or QND measurement). One can refer to the article by J.-P. Poizat and P. Grangier, Phys. Rev. Lett. 70, 271 (1993).

16 The Quantum Eraser

This chapter deals with a quantum process where the superposition of two probability amplitudes leads to an interference phenomenon. The two amplitudes can be associated with two quantum paths, as in a double slit interference experiment. We shall ﬁrst show that these interferences disappear if an intermediate measurement gives information about which path has actually been followed. Next, we shall see how interferences can actually reappear if this information is “erased” by a quantum device. We consider a beam of neutrons, which are particles of charge zero and spin 1/2, propagating along the x axis with velocity v. In all what follows, the motion of the neutrons in space is treated classically as a uniform linear motion. Only the evolution of their spin states is treated quantum mechanically.

16.1 Magnetic Resonance The eigenstates of the z component of the neutron spin are noted |n : + and |n : −. A constant uniform magnetic ﬁeld B 0 = B0 uz is applied along the z axis (uz is the unit vector along the z axis). The magnetic moment of the ˆn ˆ n , where γn is the gyromagnetic ratio and S ˆ n = γn S neutron is denoted µ the spin operator of the neutron. 16.1.1. What are the magnetic energy levels of a neutron in the presence of the ﬁeld B 0 ? Express the result in terms of ω0 = −γn B0 . 16.1.2. The neutrons cross a cavity of length L between times t0 and t1 = t0 + L/v. Inside this cavity, in addition to the constant ﬁeld B 0 , a rotating ﬁeld B 1 (t) is applied. The ﬁeld B 1 (t) lies in the (x, y) plane and it has a constant angular frequency ω: B 1 (t) = B1 (cos ωt ux + sin ωt uy ) .

(16.1)

Let |ψn (t) = α+ (t)|n : + + α− (t)|n : − be the neutron spin state at time t, and consider a neutron entering the cavity at time t0 .

156

16 The Quantum Eraser

(a) Write the equations of evolution for α± (t) when t0 ≤ t ≤ t1 . We set hereafter ω1 = −γn B1 . (b) Setting α± (t) = β± (t) exp[∓iω(t − t0 )/2], show that the problem reduces to a diﬀerential system with constant coeﬃcients. (c) We assume that we are near the resonance: |ω−ω0 | ω1 , and that terms proportional to (ω − ω0 ) may be neglected in the previous equations. Check that, within this approximation, one has, for t0 ≤ t ≤ t1 , β± (t) = β± (t0 ) cos θ − ie∓iωt0 β∓ (t0 ) sin θ , where θ = ω1 (t − t0 )/2. (d) Show that the spin state at time t1 , when the neutron leaves the cavity, can be written as:

α+ (t0 ) α+ (t1 ) = U (t0 , t1 ) (16.2) α− (t1 ) α− (t0 ) where the matrix U (t0 , t1 ) is

U (t0 , t1 ) =

e−iχ cos φ −ie−iδ sin φ −ieiδ sin φ eiχ cos φ

,

(16.3)

with φ = ω1 (t1 − t0 )/2, χ = ω(t1 − t0 )/2 and δ = ω(t1 + t0 )/2.

16.2 Ramsey Fringes The neutrons are initially in the spin state |n : −. They successively cross two identical cavities of the type described above. This is called Ramsey conﬁguration and it is shown in Fig. 16.1. The same oscillating ﬁeld B 1 (t), given

Fig. 16.1. Ramsey’s conﬁguration; the role of the detecting atom A is speciﬁed in parts 3 and 4

16.2 Ramsey Fringes

157

by 16.1, is applied in both cavities. The modulus B1 of this ﬁeld is adjusted so as to satisfy the condition φ = π/4. The constant ﬁeld B 0 is applied throughout the experimental setup. At the end of this setup, one measures the number of outgoing neutrons which have ﬂipped their spin and are in the ﬁnal state |n : +. This is done for several values of ω in the vicinity of ω = ω0 . 16.2.1. At time t0 , a neutron enters the ﬁrst cavity in the state |n : −. What is its spin state, and what is the probability to ﬁnd it in the state |n : +, when it leaves the cavity? 16.2.2. The same neutron enters the second cavity at time t0 = t1 + T , with T = D/v where D is the distance between the two cavities. Between the two cavities the spin precesses freely around B 0 . What is the spin state of the neutron at time t0 ? 16.2.3. Let t1 be the time when the neutron leaves the second cavity: t1 −t0 = t1 − t0 . Express the quantity δ = ω(t1 + t0 )/2 in terms of ω, t0 , t1 and T . Write the transition matrix U (t0 , t1 ) in the second cavity. 16.2.4. Calculate the probability P+ of detecting the neutron in the state |n : + after the second cavity. Show that it is an oscillating function of (ω0 −ω)T . Explain why this result can be interpreted as an interference process. 16.2.5. In practice, the velocities of the neutrons have some dispersion around the mean value v. This results in a dispersion in the time T to get from one cavity to the other. A typical experimental result giving the intensity of the outgoing beam in the state |n : + as a function of the frequency ν = ω/2π of the rotating ﬁeld B 1 is shown in Fig. 16.2. (a) Explain the shape of this curve by averaging the previous result over the distribution

Fig. 16.2. Intensity of the outgoing beam in the state |n : + as a function of the frequency ω/2π for a neutron beam with some velocity dispersion

158

16 The Quantum Eraser

We recall that

∞ −∞

2 2 1 dp(T ) = √ e−(T −T0 ) /2τ dT . τ 2π

cos(ΩT ) dp(T ) = e−Ω

2 2

τ /2

cos(ΩT0 ).

(b) In the above experiment, the value of the magnetic ﬁeld was B0 = 2.57× 10−2 T and the distance D = 1.6 m. Calculate the magnetic moment of the neutron. Evaluate the average velocity v0 = D/T0 and the velocity dispersion δv = v0 τ /T0 of the neutron beam. (c) Which optical interference experiment is the result reminiscent of? 16.2.6. Suppose one inserts between the two cavities of Fig. 16.1 a device which can measure the z component of the neutron spin (the principle of such a detector is presented in the next section). Determine the probability P+,+ of detecting the neutron in the state |n : + between the two cavities and in the state |n : + when it leaves the second cavity, and the probability P−,+ of detecting the neutron in the state |n : − between the cavities and in the state |n : + when it leaves the second cavity. Check that one does not have P+ = P+,+ + P−,+ and comment on this fact.

16.3 Detection of the Neutron Spin State In order to measure the spin of a neutron, one lets it interact during a time ˆ a . Let |a : ± τ with a spin 1/2 atom at rest. The atom’s spin operator is S ˆ be the two eigenstates of the observable Saz . After the interaction between the neutron and the atom, one measures the spin of the atom. Under certain conditions, as we shall see, one can deduce the spin state of the neutron after this measurement. 16.3.1. Spin States of the Atom. Let |a : ±x be the eigenstates of Sˆax and |a : ±y those of Sˆay . Write |a : ±x and |a : ±y in the basis {|a : +, |a : −}. Express |a : ±y in terms of |a : ±x. 16.3.2. We assume that the neutron–atom interaction does not aﬀect the neutron’s trajectory. We represent the interaction between the neutron and the atom by a very simple model. This interaction is assumed to last a ﬁnite time τ during which the neutron–atom interaction Hamiltonian has the form 2A ˆ Snz ⊗ Sˆax , Vˆ = h ¯

(16.4)

where A is a constant. We neglect the action of any external ﬁeld, including B 0 , during the time τ . Explain why Sˆnz and Vˆ commute. Give their common eigenstates and the corresponding eigenvalues.

16.4 A Quantum Eraser

159

16.3.3. We hereafter assume that the interaction time τ is adjusted in such a way that Aτ = π/2 . Suppose the initial state of the system is |ψ(0) = |n : + ⊗ |a : +y . Calculate the ﬁnal state of the system |ψ(τ ). Answer the same question if the initial state is |ψ(0) = |n : − ⊗ |a : +y. 16.3.4. We now suppose that the initial spin state is |ψ(0) = (α+ |n : + + α− |n : −) ⊗ |a : +y. After the neutron–atom interaction described above, one measures the z component Saz of the atom’s spin. (a) What results can one ﬁnd, and with what probabilities? (b) After this measurement, what prediction can one make about the value of the z component of the neutron spin? Is it necessary to let the neutron interact with another measuring apparatus in order to know Snz once the value of Saz is known?

16.4 A Quantum Eraser We have seen above that if one measures the spin state of the neutron between the two cavities, the interference signal disappears. In this section, we will show that it is possible to recover an interference if the information left by the neutron on the detecting atom is “erased” by an appropriate measurement. A neutron, initially in the spin state |n : −, is sent into the two-cavity system. Immediately after the ﬁrst cavity, there is a detecting atom of the type discussed above, prepared in the spin state |a : +y. By assumption, the spin state of the atom evolves only during the time interval τ when it interacts with the neutron. 16.4.1. Write the spin state of the neutron–atom system when the neutron is: (a) just leaving the ﬁrst cavity (time t1 ), before interacting with the atom; (b) just after the interaction with the atom (time t1 + τ ); (c) entering the second cavity (time t0 ); (d) just leaving the second cavity ( time t1 ). 16.4.2. What is the probability to ﬁnd the neutron in the state |n : + at time t1 ? Does this probability reﬂect an interference phenomenon? Interpret the result.

160

16 The Quantum Eraser

16.4.3. At time t1 , Bob measures the z component of the neutron spin and Alice measures the y component of the atom’s spin. Assume both measurements give +¯h/2. Show that the corresponding probability reﬂects an interference phenomenon. 16.4.4. Is this result compatible with the conclusion of question 4.2? 16.4.5. In your opinion, which of the following three statements are appropriate, and for what reasons? (a) When Alice performs a measurement on the atom, Bob sees at once an interference appear in the signal he is measuring on the neutron. (b) Knowing the result obtained by Alice on each event, Bob can select a subsample of his own events which displays an interference phenomenon. (c) The experiment corresponds to an interference between two quantum paths for the neutron spin. By restoring the initial state of the atom, the measurement done by Alice erases the information concerning which quantum path is followed by the neutron spin, and allows interferences to reappear. 16.4.6. Alice now measures the component of the atom’s spin along an arbitrary axis deﬁned by the unit vector w. Show that the contrast of the interferences varies proportionally to | sin η|, where cos η = w.uz . Interpret the result.

16.5 Solutions Section 16.1: Magnetic Resonance 16.1.1. The magnetic energy levels are: E± = ∓γn ¯hB0 /2 = ±¯ hω0 /2. 16.1.2. (a) The Hamiltonian is ¯ h H= 2

ω0 ω1 eiωt

ω1 e−iωt −ω0

.

Therefore, the evolution equations are iα˙ + =

ω0 ω1 −iωt α+ + e α− ; 2 2

iα˙ − = −

ω0 ω1 +iωt α− + e α+ . 2 2

(b) With the variables β± (t) = α± (t) exp[±iω(t − t0 )/2], we obtain iβ˙ + =

ω0 − ω ω1 −iωt0 β+ + e β− 2 2

;

iβ˙ − =

ω − ω0 ω1 iωt0 β− + e β+ . 2 2

16.5 Solutions

161

(c) If |ω0 − ω| ω1 , we have, to a good approximation, the diﬀerential system ω1 −iωt0 ω1 iωt0 e e β+ , β− ; iβ˙ − = iβ˙ + = 2 2 whose solution is indeed

β± (t) = β± (t0 ) cos

ω1 (t − t0 ) ω1 (t − t0 ) − i e∓iωt0 β∓ (t0 ) sin . 2 2

(d) Deﬁning φ = ω1 (t1 − t0 )/2 , χ = ω(t1 − t0 )/2 , δ = ω(t1 + t0 )/2, we obtain " # α+ (t1 ) = e−iχ β+ (t1 ) = e−iχ α+ (t0 ) cos φ − iα− (t0 )e−iωt0 sin φ " # α− (t1 ) = eiχ β− (t1 ) = e+iχ α− (t0 ) cos φ − iα+ (t0 )e+iωt0 sin φ , and, therefore,

U=

e−iχ cos φ −i eiδ sin φ

−i e−iδ sin φ eiχ cos φ

.

Section 16.2: Ramsey Fringes 16.2.1. We assume φ = π/4; the initial conditions are: α+ (t0 ) = 0 , α− (t0 ) = 1. At time t1 the state is

1 |ψ(t1 ) = √ −i e−iδ |n : + + eiχ |n : − . 2 √ √ In other words α+ (t1 ) = −ie−iδ / 2, α− (t1 ) = eiχ / 2, and P± = 1/2. 16.2.2. We set T = D/v. The neutron spin precesses freely between the two cavities during time T , and we obtain

1 α+ (t0 ) −ie−iδ e−iω0 T /2 =√ . (16.5) α− (t0 ) eiχ e+iω0 T /2 2 16.2.3. By deﬁnition, t0 = t1 +T and t1 = 2t1 −t0 +T , therefore the transition matrix in the second cavity is

−iχ e cos φ −ie−iδ sin φ U = −ieiδ sin φ eiχ cos φ with φ = φ = ω1 (t1 − t0 )/2, χ = χ = ω(t1 − t0 )/2. Only the parameter δ is changed into δ = ω(t1 + t0 )/2 = ω(3t1 + 2T − t0 )/2 .

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16 The Quantum Eraser

16.2.4. The probability amplitude for detecting the neutron in state |+ after the second cavity is obtained by (i) applying the matrix U to the vector (16.5), (ii) calculating the scalar product of the result with |n : +. We obtain α+ (t1 ) =

1 −i(χ+δ+ω0 T /2) − ie−i(δ −χ−ω0 T /2) . −ie 2

Since δ + χ = ωt1

δ − χ =

ω (3t1 + 2T − t0 − t1 + t0 ) = ω(t1 + T ) , 2

we have α+ (t1 ) = −

i −iω(t1 +T /2) −i(ω0 −ω)T /2 e + ei(ω0 −ω)T /2 . e 2

(16.6)

Therefore, the probability that the neutron spin has ﬂipped in the two-cavity system is (ω − ω0 )T . P+ = |α+ (t1 )|2 = cos2 2 With the approximation of Sect. 1.2c, the probability for a spin ﬂip in a single cavity is independent of ω, and is equal to 1/2. In contrast, the present result for two cavities exhibits a strong modulation of the spin ﬂip probability, between 1 (e.g. for ω = ω0 ) and 0 (e.g. for (ω − ω0 )T = π). This modulation results from an interference process of the two quantum paths corresponding respectively to: • a spin ﬂip in the ﬁrst cavity, and no ﬂip in the second one, • no ﬂip in the ﬁrst cavity and a spin ﬂip in the second one. Each of these paths has a probability 1/2, so that the sum of the probability amplitudes (16.6) is fully modulated. 16.2.5. (a) Since cos2 φ/2 = (1 + cos φ)/2, the averaged probability distribution is cos2

(ω − ω0 )T 2

=

# " 1 1 −(ω−ω0 )2 τ 2 /2 + e cos (ω − ω0 )T0 . 2 2

(16.7)

This form agrees with the observed variation in ω of the experimental signal. The central maximum, which is located at ω/2π = 748.8 kHz corresponds to ω = ω0 . For that value, a constructive interference appears whatever the neutron velocity. The lateral maxima and minima are less peaked, however, since the position of a lateral peak is velocity dependent. The ﬁrst two lateral maxima correspond to (ω − ω0 )T0 ±2π. Their amplitude is reduced, compared to the central peak, by a factor exp(−2π 2 τ 2 /T02 ). (b) The angular frequency ω0 is related to the magnetic moment of the neutron by h ¯ ω0 = 2µn B0 which leads to µn = 9.65 × 10−27 J T−1 . The time T0 can be deduced from the spacing between the central maximum and a

16.5 Solutions

163

lateral one. The ﬁrst lateral maximum occurs at 0.77 kHz from the resonance, hence T0 = 1.3 ms. This corresponds to an average velocity v0 = 1230 m s−1 . The ratio of intensities between the second lateral maximum and the central one is roughly 0.55. This is approximately equal to exp(−8π 2 τ 2 /T02 ), and gives τ /T0 ≈ 0.087, and δv ≈ 110 m s−1 . (c) This experiment can be compared to a Young double slit interference experiment with polychromatic light.The central fringe (corresponding to the peak at ω = ω0 ) remains bright, but the contrast of the interferences decreases rapidly as one departs from the center. In fact, the maxima for some frequencies correspond to minima for others. 16.2.6. The probability P++ is the product of the two probabilities: the probability to ﬁnd the neutron in the state |n : + when it leaves the ﬁrst cavity (p = 1/2) and, knowing that it is in the state |n : +, the probability to ﬁnd it in the same state when it leaves the second cavity (p = 1/2); this gives P+,+ = 1/4. Similarly P−,+ = 1/4. The sum P+,+ + P−,+ = 1/2 does not display any interference, since one has measured in which cavity the neutron spin has ﬂipped. This is very similar to an electron double-slit interference experiment if one measures which slit the electron goes through. Section 16.3: Detection of the Neutron Spin State 16.3.1. By deﬁnition: 1 |a : ±x = √ (|a : + ± |a : −) 2 1 |a : ±y = √ (|a : + ± i|a : −) 2 and these states are related to one another by |a : ±y =

1 ((1 ± i)|a : +x + (1 ∓ i)|a : −x) . 2

16.3.2. The operators Sˆnz and Sˆax commute since they act in two diﬀerent Hilbert spaces; therefore [Sˆnz , Vˆ ] = 0. The common eigenvectors of Sˆnz and Vˆ , and the corresponding eigenvalues are h/2 V = ±A¯ h/2 , |n : + ⊗ |a : ±x Snz = +¯ h/2 V = ∓A¯ h/2 . |n : − ⊗ |a : ±x Snz = −¯ The operators Sˆnz and Vˆ form a complete set of commuting operators as far as spin variables are concerned. 16.3.3. Expanding in terms of the energy eigenstates, one obtains for |ψ(0) = |n : + ⊗ |a : +y : 1 |ψ(τ ) = |n : + ⊗ (1 + i)e−iAτ /2 |a : +x + (1 − i)eiAτ /2 |a : −x , 2

164

16 The Quantum Eraser

i.e. for Aτ /2 = π/4: 1 |ψ(τ ) = √ |n : + ⊗ (|a : +x + |a : −x) 2 = |n : + ⊗ |a : + . Similarly, if |ψ(0) = |n : − ⊗ |a : +y, then |ψ(τ ) = i|n : − ⊗ |a : −. Physically, this means that the neutron’s spin state does not change since it is an eigenstate of Vˆ , while the atom’s spin precesses around the x axis with angular frequency A. At time τ = π/(2A), it lies along the z axis. 16.3.4. If the initial state is |ψ(0) = (α+ |n : + + α− |n : −) ⊗ |a : +y, the state after the interaction is |ψ(τ ) = α+ |n : + ⊗ |a : + + iα− |n : − ⊗ |a : − . The measurement of the z component of the atom’s spin gives +¯ h/2, with probability |α+ |2 and state |n : + ⊗ |a : + after the measurement, or −¯ h/2 with probability |α− |2 and state |n : − ⊗ |a : − after the measurement. In both cases, after measuring the z component of the atom’s spin, the neutron spin state is known: it is the same as that of the measured atom. It is not necessary to let the neutron interact with another measuring apparatus in order to know the value of Snz . Section 16.4: A Quantum Eraser 16.4.1. The successive states are: step (a) step (b) step (c)

1 √ −ie−iδ |n : + ⊗ |a : +y + eiχ |n : − ⊗ |a : +y 2

1 −iδ √ −ie |n : + ⊗ |a : + + ieiχ |n : − ⊗ |a : − 2 1 √ − ie−i(δ+ω0 T /2) |n : + ⊗ |a : + 2 +iei(χ+ω0 T /2) |n : − ⊗ |a : − .

Finally, when the neutron leaves the second cavity (step d), the state of the system is: |ψf =

1 − ie−i(δ+ω0 T /2) e−iχ |n : + − ieiδ |n : − ⊗ |a : + 2

+iei(χ+ω0 T /2) −ie−iδ |n : + + eiχ |n : − ⊗ |a : − .

16.4.2. The probability to ﬁnd the neutron in state |+ is the sum of the probabilities for ﬁnding:

16.5 Solutions

165

(a) the neutron in state + and the atom in state +, i.e. the square of the modulus of the coeﬃcient of |n : + ⊗ |a : + (1/4 in the present case), (b) the neutron in state + and the atom in state − (probability 1/4 again). One gets therefore P+ = 1/4 + 1/4 = 1/2: There are no interferences since the quantum path leading in the end to a spin ﬂip of the neutron can be determined from the state of the atom. 16.4.3. One can expand the vectors |a : ± on |a : ±y: 1 |ψf = √ − ie−i(δ+ω0 T /2) e−iχ |n : + − ieiδ |n : − 2 2 ⊗ (|a : +y + |a : −y) i(χ+ω0 T /2) +e −ie−iδ |n : + + eiχ |n : − ⊗ (|a : +y − |a : −y) The probability amplitude that Bob ﬁnds +¯ h/2 along the z axis while Alice ﬁnds +¯ h/2 along the y axis is the coeﬃcient of the term |n : + ⊗ |a : +y in the above expansion. Equivalently, the probability is obtained by projecting the state onto |n : + ⊗ |a : +y, and squaring. One obtains

2 ¯h h ¯ 1 P Snz = , Say = = −ie−i(δ+χ+ω0 T /2) − iei(χ−δ +ω0 T /2) 2 2 8 (ω − ω0 )T 1 , = cos2 2 2 which clearly exhibits a modulation reﬂecting an interference phenomenon. Similarly, one ﬁnds that

(ω − ω0 )T h ¯ h ¯ 1 , P Snz = , Say = − = sin2 2 2 2 2 which is also modulated. 16.4.4. This result is compatible with the result 4.2. Indeed the sum of the two probabilities calculated above is 1/2 as in 4.2. If Bob does not know the result found by Alice, or if Alice does not perform a measurement, which is equivalent from his point of view, Bob sees no interferences. The interferences only arise for the joint probability P (Snz , Say ). 16.4.5. (a) This ﬁrst statement is obviously wrong. As seen in question 4.2, if the atom A is present, Bob no longer sees oscillations (in ω − ω0 ) of the probability for detecting the neutron in the state |+. This probability is equal to 1/2 whatever Alice does. Notice that if the statement were correct, this would imply instantaneous transmission of information from Alice to Bob. By seeing interferences appear, Bob would know immediately that Alice is performing an experiment, even though she may be very far away.

166

16 The Quantum Eraser

(b) This second statement is correct. If Alice and Bob put together all their results, and if they select the subsample of events for which Alice ﬁnds +¯ h/2, then the number of events for which Bob also ﬁnds +¯h/2 varies like cos2 ((ω − ω0 )T /2); they recover interferences for this subset of events. In the complementary set, where Alice has found −¯ h/2, the number of Bob’s results giving +¯ h/2 varies like sin2 ((ω − ω0 )T /2). This search for correlations between events occurring in diﬀerent detectors is a common procedure, in particle physics for instance. (c) This third statement, although less precise but more picturesque than the previous one, is nevertheless acceptable. The cos2 ((ω − ω0 )T /2) signal found in Sect. 2 can be interpreted as the interference of the amplitudes corresponding to two quantum paths for the neutron spin which is initially in the state |n : −; either its spin ﬂips in the ﬁrst cavity, or it ﬂips in the second one. If there exists a possibility to determine which quantum path is followed by the system, interferences cannot appear. It is necessary to “erase” this information, which is carried by the atom, in order to observe “some” interferences. After Alice has measured the atom’s spin along the y axis, she has, in some sense “restored” the initial state of the system, and this enables Bob to see some interferences. It is questionable to say that information has been erased: one may feel that, on the contrary, extra information has been acquired. Notice that the statement in the text does not specify in which physical quantity the interferences appear. Notice also that the order of the measurements made by Alice and Bob has no importance, contrary to what this third statement seems to imply. 16.4.6. Alice can measure along the axis w = sin η uy +cos η uz , in the (y, z) plane, for instance. Projecting |ψf onto the eigenstate of Sˆaw with eigenvalue +¯ h/2, i.e. cos(η/2) |a : −,

# a calculation similar to 4.3 leads " |a : + + i sin(η/2) to a probability 1 + sin η cos (ω − ω0 )T /2. If η = 0 or π (measurement along the z axis) there are no interferences. For η = π/2 and 3π/2 or, more generally, if Alice measures in the (x, y) plane, the contrast of the interferences, | sin η|, is maximum.

16.6 Comments Ramsey Fringes with Neutrons. The experimental curve given in the text is taken from J.H. Smith et al., Phys. Rev. 108, 120 (1957). Since then, the technique of Ramsey fringes has been considerably improved. Nowadays one proceeds diﬀerently. One stores neutrons in a “bottle” for a time of the order of 100 s and applies two radiofrequency pulses at the begining and at the end of the storage. The elapsed time between the two pulses is 70 s, compared to 1.3 ms here. This improves enormously the accuracy of the frequency measurement. Such experiments are actually devised to measure the electric dipole moment of the neutron, of fundamental interest in relation to time-reversal

16.6 Comments

167

invariance. They set a very small upper bound on this quantity (K.F. Smith et al., Phys. Lett. 234, 191 (1990)). Non Destructive Quantum Measurements. The structure of the interaction Hamiltonian considered in the text has been chosen in order to provide a simple description of the quantum eraser eﬀect. Realistic examples of nondestructive quantum measurements can be found in J.P. Poizat and P. Grangier, Phys. Rev. Lett. 70, 271 (1993), and S.M. Barnett, Nature, Vol. 362, p. 113, March 1993.

17 A Quantum Thermometer

We study here the measurement of the cyclotron motion of an electron. The particle is conﬁned in a Penning trap and it is coupled to the thermal radiation which causes quantum transitions of the system between various energy levels. In all the chapter, we neglect spin eﬀects. The method and results come from an experiment performed at Harvard University in 1999. We consider an electron of mass M and charge q (q < 0), conﬁned in a Penning trap. This trap consists in the superposition of a uniform magnetic ﬁeld B = Bez (B > 0) and an electric ﬁeld which derives from the potential Φ(r) whose power expansion near the origin is: Φ(r) =

M ωz2 2 2z − x2 − y 2 . 4q

(17.1)

The positive quantity ωz has the dimension of an angular frequency. In all this chapter we set ωc = |q|B/M (ωc is called the cyclotron angular frequency) and we assume that ωz ωc . Useful constants: M = 9.1 10−31 kg ; q = −1.6 10−19 C; h = 6.63 10−34 J s; Boltzmann’s constant kB = 1.38 10−23 J K−1 .

17.1 The Penning Trap in Classical Mechanics We recall that the Lorentz force acting on a charged particle moving in an electromagnetic ﬁeld is F = q(E + v × B). 17.1.1. Check that Φ(r) satisﬁes the Laplace equation ∆Φ = 0. What is the shape of a surface of constant potential Φ(r) =Const? 17.1.2. Show that the classical equation of motion of the electron in the trap is: x ¨ + ωc y˙ −

ωz2 x=0 2

y¨ − ωc x˙ −

ωz2 y=0 2

z¨ + ωz2 z = 0 .

170

17 A Quantum Thermometer

17.1.3. What is the type of motion along the z axis? 17.1.4. In order to study the component of the motion in the xy plane, we set α = x + iy. (a) What is the diﬀerential equation satisﬁed by α(t)? (b) We seek a solution of this equation of the form α(t) = α0 eiωt . Show that ω is a solution of the equation: ω 2 − ωc ω +

ωz2 =0. 2

(c) We note ωr and ωl the two roots of this equation with ωr > ωl . Show that: ω2 ωl z . ωr ωc 2ωc 17.1.5. We consider the values B = 5.3 T and ωz /(2π) = 64 MHz. (a) Show that the most general motion of the electron in the Penning trap is the superposition of three harmonic oscillator motions. (b) Calculate the frequencies of these motions. (c) Draw the projection on the xy plane of the classical trajectory of the trapped electron, assuming that αr αl (the positive quantities αr and αl represent the amplitudes of the motions of angular frequencies ωr and ωl ).

17.2 The Penning Trap in Quantum Mechanics ˆ the position and momentum operators of the electron. The We note rˆ and p Hamiltonian of the electron in the Penning trap is, neglecting spin eﬀects: 2 ˆ = 1 (ˆ H p − qA(ˆ r )) + qΦ(ˆ r) , 2M

where the electrostatic potential Φ(r) is given by (17.1). For the magnetic vector potential, we choose the form A(r) = B × r/2. ˆ = 17.2.1. Expand the Hamiltonian and show that it can be written as H ˆ xy + H ˆ z , where H ˆ xy only involves the operators x ˆz H ˆ, yˆ, pˆx and pˆy , while H only involves the operators zˆ and pˆz . ˆ z possess a common eigenbasis? ˆ xy and H Do H 17.2.2. We are now interested in the motion along the z axis. This is called the axial motion. Recall without giving any proof: ˆ z in the ˆ†z which allow to write H (a) the expression of the operators a ˆz and a † † ˆ ˆ ˆ ¯ ωz (Nz + 1/2) with Nz = a ˆz a ˆz and [ˆ az , a ˆz ] = 1; form Hz = h ˆ ˆ (b) the eigenvalues of Nz and Hz .

17.2 The Penning Trap in Quantum Mechanics

171

17.2.3. We now consider the motion in the xy plane under the eﬀect of the ˆ xy . We set Ω = ωc2 − 2ωz2 /2. We introduce the right and left Hamiltonian H ˆl : annihilation operators a ˆr and a i MΩ (ˆ x − iˆ y) + √ (ˆ px − iˆ py ) a ˆr = 4¯ h 4¯ hM Ω i MΩ (ˆ x + iˆ y) + √ (ˆ px + iˆ a ˆl = py ) . 4¯ h 4¯ hM Ω (a) Show that [ˆ ar , a ˆ†r ] = [ˆ al , a ˆ†l ] = 1. (b) Show that any left operator commutes with any right operator, i.e.: ˆl ] = 0 [ˆ ar , a

[ˆ ar , a ˆ†l ] = 0

[ˆ a†r , a ˆl ] = 0

[ˆ a†r , a ˆ†l ] = 0 .

(c) Recall the eigenvalues of n ˆr = a ˆ†r a ˆr and n ˆl = a ˆ†l a ˆl (no proof is required). ˆ l possess a common eigenbasis? Do n ˆ r and n ˆ xy can be written as: (d) Show that the Hamiltonian H ˆ xy = h H ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) , where the angular frequencies ωr and ωl have been introduced in Sect. 17.1. ˆ xy . (e) Deduce from this the eigenvalues of the Hamiltonian H 17.2.4. We note |ψ(t) the state of the system at time t and we deﬁne ar |ψ(t) and al (t) = ψ(t)|ˆ al |ψ(t). Using the Ehrenfest thear (t) = ψ(t)|ˆ orem, calculate dar /dt and dal /dt. Integrate these equations and calculate the expectation value of the electron’s position (x(t), y(t)) in the xy plane. We set ar (0) = ρr e−iφr and al (0) = ρl eiφl , where ρr and ρl are real and positive. Show that the time evolution of the expectation value of the electron position r(t) is similar to the classical evolution found in Sect. 17.1. ˆ corresponding to the eigenvalues 0 17.2.5. We note |φ0 the eigenstate of H ˆ ˆ l and Nz . for each of the operators n ˆr , n (a) Determine the corresponding wave function φ0 (r) (it is not necessary to normalize the result). (b) Using the same numerical values as in question 17.1.5, evaluate the spatial extension of φ0 (r). 17.2.6. The experiment is performed at temperatures T ranging between 0.1 K and 4 K. Compare the characteristic thermal energy kB T to each of the energy quanta of the “cyclotron”, “axial” and “magnetron” motions (associˆz and n ˆ l ). For which of these motions does the ated respectively with n ˆr , N discrete nature of the energy spectrum play an important role?

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17 A Quantum Thermometer

17.3 Coupling of the Cyclotron and Axial Motions We now study a method for detecting the cyclotron motion. This method uses a small coupling between this motion and the axial motion. The coupling is produced by an inhomogeneous magnetic ﬁeld, and it can be described by the additional term in the Hamiltonian: ˆ = M ω2 n ˆ2 . W z ˆr z 2 The experimental conditions are chosen such that = 4 × 10−7 . ˆ +W ˆ using the operators n ˆc = H ˆr , 17.3.1. Write the total Hamiltonian H n ˆ l , pˆz and zˆ. 17.3.2. Show that the excitation numbers of the cyclotron motion (ˆ nr ) and of the magnetron motion (ˆ nl ) are constants of the motion. 17.3.3. Consider the eigensubspace Enr ,nl of n ˆ r and n ˆ l , corresponding to the eigenvalues nr and nl . ˆ c in this subspace. (a) Write the form of H (b) Show that the axial motion is harmonic if the system is prepared in a state belonging to Enr ,nl . Give its frequency in terms of nr and nl . ˆ c inside En ,n . (c) Give the eigenvalues and eigenstates of H r

l

ˆ c can be 17.3.4. Deduce from the previous question that the eigenstates of H labeled by 3 quantum numbers, nr , nl , nz . We write these states as |nr , nl , nz . Give the energy eigenvalues in terms of these quantum numbers and of ωr , ωl , ωz and . 17.3.5. One measures the beat between a highly stable oscillator of frequency ωz /(2π) (delivering a signal proportional to sin(ωz t)) and the current induced in an electric circuit by the axial motion. This latter current is proportional to pz (t). (a) Calculate the time evolution of the expectation values of the position and momentum operators zˆ and pˆz assuming that the state of the electron is restricted to be in the subspace Enr ,nl . We choose the initial conditions ˆ z (t = 0) = z0 and pˆz (t = 0) = 0. (b) To ﬁrst order in , what is the phase diﬀerence ϕ between the detected current and the stable oscillator after a time τ ? Show that the measurement of this phase diﬀerence provides a measurement of the excitation number of the cyclotron motion. 17.3.6. We now assume that the electron is in an arbitrary state cnr ,nl ,nz |nr , nl , nz . |Ψ = nr ,nl ,nz

(a) We measure the phase diﬀerence ϕ on a time interval ranging from t = 0 to t = τ . What are the possible results ϕk of the measurement? Show that

17.4 A Quantum Thermometer

173

this provides a means to determine the excitation number of the cyclotron motion. (b) What is the state of the electron after a measurement giving the result ϕk ? (c) We choose τ = 0.1 s and we assume that the measurement of ϕ is done with an accuracy of π/10. Using the previous values of the physical parameters, show that this accuracy leads to an unambiguous determination of the cyclotron excitation number. (d) After a measurement giving the result ϕk , we let the system evolve for a ˆ c . We then perform length of time T under the action of the Hamiltonian H a new measurement of ϕ. What results do we expect?

17.4 A Quantum Thermometer In practice, the cyclotron motion is in thermal equilibrium with a thermostat at temperature T . We recall that, in that situation, the thermal ﬂuctuations can excite the system in an energy level En with some probability pn . We perform successive measurements of the phase diﬀerence ϕ in the time intervals [0, τ ], [τ, 2τ ], . . ., [(N − 1)τ, N τ ]. The total duration N τ of this series of measurements for a given temperature T is N τ = 3000 seconds, i.e. a total number of results N = 3 × 104 for τ = 0.1 s. By this procedure, one can follow the variation of nr during the time interval N τ , with a time resolution of τ . 17.4.1. Two recordings of this measurement are represented on ﬁgure 17.1 for two diﬀerent temperatures. Comment on these results and explain in particular: (a) to what phenomenon are associated the sudden changes of the signal; (b) what is the fraction of the time during which the electron is in the levels nr = 0, nr = 1, nr = 2, . . . (the accuracy of a usual graduated ruler is suﬃcient). 17.4.2. The probability pn for a system to be in the energy level En is given by the Boltzmann factor pn = N exp(−En /kB T ), where N is a normalization factor. Show that for a one-dimensional harmonic oscillator the ratio pn+1 /pn does not depend on n. 17.4.3. Give an estimate of the two temperatures corresponding to the two recordings of Fig. 17.1. 17.4.4. Fig. 17.2 represents more accurate measurements of the occupation probabilities of the various cyclotron levels for several temperatures of the cryostat which contains the Penning trap. (a) Determine the normalization factor N of the probability law pn for a one-dimensional harmonic oscillator of angular frequency ω in thermal equilibrium with a thermostat at temperature T , and calculate the average excitation number n ¯ . It is convenient to set γ = h ¯ ω/(kB T ).

174

17 A Quantum Thermometer

2 1 cyclotron 0 quantum number 2 1 0 0

10

30

20

40

50

time (minutes) Fig. 17.1. Time evolution of the quantum number nr corresponding to the cyclotron motion for two temperatures Ta and Tb

(b) Justify the aspect of the curves of ﬁgure 17.2 and evaluate the corresponding temperatures of the measurements. (c) What is the order of magnitude of the lowest temperature one can measure with such a device? (d) How can one improve the sensitivity of such a “quantum thermometer”?

Occupation probability of the cyclotron level nd

1

0,1

0,01

0

1

2

Cyclotron quantum number nd

Fig. 17.2. Occupation probabilities of the energy states of the cyclotron motion. Each straight line corresponds to a given temperature (the vertical scale is logarithmic)

17.5 Solutions Section 17.1: The Penning Trap in Classical Mechanics 17.1.1. The electric ﬁeld E = −∇Φ is:

17.5 Solutions

⎛ E(r) =

M ωz2 ⎝ 2q

175

⎞

x y⎠ −2z

(17.2)

M ω2

Therefore, one has ∆Φ = −∇ · E = − 2q z (1 + 1 − 2) = 0. The potential satisﬁes Laplace’s equation in the vacuum. The surfaces of constant potential are hyperboloids of axis z. 17.1.2. Using the expression (17.2) for the electric ﬁeld, the equation of motion is: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ x x˙ x ¨ 0 2 M ωz ⎝ y ⎠ + q ⎝ y˙ ⎠ × ⎝ 0 ⎠ , M ⎝ y¨ ⎠ = 2 −2z z˙ z¨ B or, by setting ωc = −qB/M : x ¨ + ωc y˙ −

ωz2 x=0 2

y¨ − ωc x˙ −

ωz2 y=0 2

z¨ + ωz2 z = 0 .

17.1.3. Along the z axis, the motion is harmonic, of angular frequency ωz . 17.1.4. (a) The diﬀerential equation satisﬁed by α(t) is: α ¨ − iωc α˙ −

ωz2 α=0 2

(b) If we search a solution of the form α0 eiωt , we ﬁnd that ω is given by the equation: ω2 −ω 2 + ωc ω − z = 0 2 (c) The roots of this equation are: 1 1 ωr = ωl = ωc + ωc2 − 2ωz2 ωc − ωc2 − 2ωz2 2 2

2 We assume ωz ωc , i.e. ωc2 − 2ωz2 ωc 1 − ωz /ωc2 . The two roots ωr and ωl are given approximately by: ω r ωc

ωl

ωz2 . 2ωc

17.1.5. (a) We have seen previously that the motion along z is harmonic of angular frequency ωz . In order to obtain the motion in the xy plane, we integrate the equation of motion of α: α(t) = αl ei(ωl t+φl ) + αr ei(ωr t+φr ) , where αl and αr are two positive real numbers, and φl and φr are two a priori arbitrary phases. This gives the forms of x(t) and y(t) by taking the real and imaginary parts of this expression:

176

17 A Quantum Thermometer

x(t) = αl cos(ωl t + φl ) + αr cos(ωr t + φr ) y(t) = αl sin(ωl t + φl ) + αr sin(ωr t + φr ) . The motion in the xy plane is a superposition of two harmonic motions of angular frequencies ωr and ωl . (b) One ﬁnds ωc /2π = 1.48 × 1011 Hz. The frequencies of the three motions are therefore: ωl /2π 14 kHz

ωz /2π = 64 MHz

ωr /2π 150 GHz

(c) The motion in the xy plane is the superposition of two circular motions, one is of radius αr and has a high frequency (ωr ), and the other is of radius αl and has a much lower frequency (ωl ). Assuming αr αl , this results in a trajectory of the type represented on the following ﬁgure

Section 17.2: The Penning Trap in Quantum Mechanics ˆ z with: ˆ =H ˆ xy + H 17.2.1. The expansion of the Hamiltonian gives H 2 pˆ2 ωc ˆ 2 ˆ xy = pˆx + y + M (ω 2 − 2ω 2 )(ˆ H ˆ2 ) + L z z x +y 2M 2M 8 c 2

2 ˆ z = pˆz + 1 M ω 2 zˆ2 , H z 2M 2 ˆz = where we have introduced the z component of the angular momentum L ˆ xy involves only the operators x ˆ, yˆ and pˆx , pˆy , and since x ˆpˆy − yˆpˆx . Since H ˆ z only involves zˆ, pˆz , the two operators H ˆ xy and H ˆ z commute, and each of H ˆ them commutes with the total Hamiltonian H:

ˆz] = 0 ˆ xy , H [H

ˆ xy , H] ˆ =0 [H

ˆ z , H] ˆ =0. [H

ˆ in the form of a common We can therefore search for an eigenbasis of H ˆz. ˆ xy and H eigenbasis of H ˆ z corresponds to a harmonic motion of an17.2.2. (a) The Hamiltonian H gular frequency ωz . Setting pˆ M ωz ˆz = a a ˆz = x ˆ + i√ and N ˆ†z a ˆz , 2¯ h 2M ¯ hωz ˆz = one easily ﬁnds its spectrum since the Hamiltonian can be written as H ˆ hωz (Nz + 1/2). ¯

17.5 Solutions

177

(b) From the commutation relation [ˆ az , a ˆ†z ] = 1, we deduce that the eigenvalˆ z are therefore ˆ ues of Nz are the non-negative integers nz . The eigenvalues of H of the form ¯hωz (nz + 1/2). 17.2.3. We ﬁrst calculate the generic commutator: M Ω i C= (ˆ x − iη yˆ) + ξ √ (ˆ px − iη pˆy ) , 4¯ h 4¯ hM Ω MΩ i (ˆ x + iη yˆ) − ξ √ (ˆ px + iη pˆy ) 4¯ h 4¯ hM Ω where the four numbers η, ξ, η and ξ are equal to ±1. Using [ˆ x, pˆx ] = [ˆ y , pˆy ] = i¯ h, one ﬁnds: 1 C = (ξ + ξ ) (1 + ηη ) . 4 ˆ†r ] corresponds to η = η = +1 and ξ = ξ = 1, (a) The commutator [ˆ ar , a therefore [ˆ ar , a ˆ†r ] = 1. Similarly, one obtains [ˆ al , a ˆ†l ] = 1 from η = η = −1 and ξ = ξ = 1. (b) The commutator [ˆ ar , a ˆl ] corresponds to ξ = 1 and ξ = −1, hence [ˆ ar , a ˆl ] = 0. Similarly, [ˆ ar , a ˆ†l ] vanishes since it corresponds to η = −η = 1. ˆl ] and [ˆ a†r , a ˆ†l ]) also vanish if we take Therefore the other commutators ([ˆ a†r , a the Hermitian conjugates of the previous commutators. ˆ†r ] = 1 results in the fact that the eigen(c) The commutation relation [ˆ ar , a ˆl. values of n ˆ r are the non-negative integers, and the same holds for n (d) The operators n ˆ r and n ˆ l are expanded as: n ˆ r,l =

ˆz pˆ2x + pˆ2y MΩ 2 1 L + (ˆ x + yˆ2 ) − ± 4¯ hM Ω 4¯ h 2 2¯h

ˆ l ). The sum and diﬀerence where the + (resp.−) sign corresponds to n ˆ r (resp. n of the roots of the equation ω 2 − ωc ω + ωz2 /2 = 0 are: ωr − ωl = ωc2 − 2ωz2 = 2Ω . ωr + ω l = ωc We obtain the anticipated result: ˆ xy = h H ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) ˆ xy can therefore be labeled by the two (non(e) The eigenvectors of H negative) integer quantum numbers nr and nl corresponding to the eigenvalues ˆ l . The corresponding eigenstates are noted |nr , nl . The eigenvalue of n ˆ r and n ˆ xy associated to the vector |nr , nl is ¯hωr (nr + 1/2) − ¯hωl (nl + 1/2). of H 17.2.4. We have: ˆ =h [ˆ ar , H] ¯ ωr [ˆ ar , a ˆ†r a ˆr ] = h ¯ ωr a ˆr .

178

17 A Quantum Thermometer

Owing to Ehrenfest theorem we therefore ﬁnd: a˙ r = −iωr ar and a˙ l = +iωl al . The solutions of these two equations are: ar (t) = ar (0) e−iωr t

al (t) = al (0) e+iωl t

The expectation value of the position in the xy plane can be calculated by using: i i MΩ MΩ x+ √ y+ √ px py . i(ar −al ) = ar +al = h ¯ h ¯ hM Ω ¯ ¯hM Ω In other words: h ¯ x(t) = Re (ar (t) + al (t)) MΩ

y(t) =

¯ h Re (iar (t) − ial (t)) . MΩ

Setting ar (0) = ρr e−iφr and al (0) = ρl eiφl , we obtain: h ¯ (ρr cos(ωr t + φr ) + ρl cos(ωl t + φl )) x(t) = MΩ h ¯ (ρr sin(ωr t + φr ) + ρl sin(ωl t + φl )) y(t) = MΩ As in the classical motion, the coordinates x and y are the sums of two sinusoidal functions of angular frequencies ωr and ωl . The x and y components have equal amplitudes and are phase shifted by π/2 with respect to each other. The average motion in the xy plane is therefore the superposition of two uniform circular motions of angular frequencies ωr and ωl . The trajectory is the same as in question 1.5. 17.2.5. (a) The wave function φ0 (r) corresponding to nr = nl = nz = 0 can be written as the product of three functions in the variables x + iy, x − iy and z. This function must satisfy a ˆµ φ0 (r) = 0, with µ = r, l, z. Therefore, ˆl respectively: setting η = + and η = − for a ˆr and a

∂ ∂ MΩ M ωz ∂ − iη + (x − iηy) φ0 (r) = 0 + z φ0 (r) = 0 . ∂x ∂y h ¯ ∂z ¯h By adding and subtracting the two equation for η = ±, we obtain:

∂ MΩ MΩ ∂ + x φ0 (r) = 0 + y φ0 (r) = 0 ∂x h ¯ ∂y ¯h The function φ0 (r) is therefore a product of three gaussian functions in the variables x, y, z: 2 2 2 2 2 φ0 (r) ∝ e−(x +y )/4r0 e−z /4z0

with r0 =

¯ h 2M Ω

and

z0 =

¯h . 2M ωz

17.5 Solutions

179

(b) The probability distribution |φ0 (r)|2 is centered at r = 0. Its extension along the x and y axes is ∆x = ∆y = r0 11 nm. Along the z axis, we have ∆z = z0 380 nm. 17.2.6. For the range of temperatures of interest the ratio kB T /¯hωµ for µ = r, l, z is kB T = 0.014 − 0.6 hωr ¯

kB T = 1.5 × 105 − 6 × 106 hωl ¯

kB T = 30 − 1300 ¯hωz

The discrete nature of the energy spectrum will play a decisive role only for ˆ†r ). Only the ﬁrst three levels of the cyclotron motion (corresponding to a ˆr , a this motion nr = 0, 1, 2, 3 will be occupied signiﬁcantly in this very low temperature domain. For the other components of the motion, of much lower frequencies than the cyclotron motion, one expects that the thermal ﬂuctuations will populate a large number of levels. The “quantum” character of the motions will be hidden under the thermal noise. Section 17.3: Coupling of the Cyclotron and Axial Motions ˆc 17.3.1. In the presence of the axial-cyclotron coupling, the Hamiltonian H is: M ωz2 pˆ2 ˆc = h H (1 + ˆ nr ) zˆ2 . ¯ ωr (ˆ nr + 1/2) − ¯ hωl (ˆ nl + 1/2) + z + 2M 2 ˆ c . The 17.3.2. It is straightforward to check that n ˆ r and n ˆ l commute with H corresponding physical quantities (i.e. the excitation numbers of the cyclotron and magnetron motions) are therefore constants of the motion. ˆ c involves only 17.3.3. (a) Inside the subspace Enr ,nl , the Hamiltonian H the operators zˆ and pˆz : 2 2 ˆ (nr ,nl ) = pˆz + M ωz (1 + nr ) zˆ2 + En ,n H c r l 2M 2

with Enr ,nl = h ¯ ωr (nr + 1/2) − ¯ hωl (nl + 1/2). (b) If the system is prepared in a state belonging to the subspace Enr ,nl , it will remain in this subspace since nr and nl are constants of the motion. Its ˆ (nr ,nl ) , corresponding to a harmonic motion is described by the Hamiltonian √ Hc oscillator along z, with frequency ωz 1 + nr . ˆ c(nr ,nl ) are ¯hωz (nz + 1/2) √1 + nr + En ,n . The (c) The eigenvalues of H r l corresponding eigenstates are the Hermite functions ψn (Z), where: h. Z = z M ωz (1 + nr )1/2 /¯

180

17 A Quantum Thermometer

17.3.4. We can perform the same operation as above inside each subspace ˆ c which we note ˆ l . We thus obtain a basis of eigenstates of H of n ˆ r and n |nr , nl , nz . The eigenvalue corresponding to each eigenstate is: √ Enr ,nl ,nz = h ¯ ωr (nr + 1/2) − ¯ hωl (nl + 1/2) + h ¯ ωz (nz + 1/2) 1 + nr Contrary to the result of Sect. 2, this basis no longer corresponds to factorized functions of the variables x ± iy and z. The coupling between the axial and cyclotron motions induces a correlation between the axial frequency and the state of the cyclotron motion. 17.3.5. (a) If the system is prepared in the subspace Enr ,nl , the axial √ Hamiltonian corresponds to a harmonic oscillator of angular frequency ωz 1 + nr . The Ehrenfest theorem then gives:

√ z(t) = z0 cos ωz t 1 + nr

√ √ pz (t) = −M ωz z0 1 + nr sin ωz t 1 + nr since the evolution equations of the quantum expectation values coincide with the classical equations for a harmonic oscillator. (b) The phase shift accumulated during a time interval τ between the detected current, proportional to pz (t), and the external oscillator is: √ ϕ = ωz τ 1 + nr − ωz τ ωz τ nr . 2 Knowing the time τ , the frequency ωz and the coupling constant , one can deduce the cyclotron excitation number nr . 17.3.6. (a) The possible results of a measurement are the numbers ϕk =

ωz τ k , 2

where k = nr is a non-negative integer. A given experimental result determines unambiguously the excitation number of the cyclotron motion. (b) The measurement postulate of quantum mechanics implies that the state |Ψ of the system after a measurement corresponds to the projection of the state vector before the measurement |Ψ on the eigensubspace corresponding to the measured result: cn(0) ,n ,nz |n(0) |Ψ ∝ r , nl , nz nl ,nz

(0)

r

l

where the integer nr corresponds to the result of the measurement of ϕ. One must further normalize the right hand side in order to obtain the state vector |Ψ .

17.5 Solutions

181

(c) For the values of the parameters given in the text, one ﬁnds φ1 = ωz τ /2 2π × 1.28. The accuracy of π/10 is much smaller than the difference between the phase shifts corresponding to nr and nr + 1, and one can indeed measure unambiguously the excitation numbers nr = 0, 1, 2, . . .. ˆr. (d) A measurement of nr prepares the system in an eigensubspace of n ˆ c , nr is a constant of the motion. Since n ˆ r commutes with the Hamiltonian H Any further measurement of the cyclotron excitation number will therefore give the same result nr , corresponding to the same phase shift ϕk . Of course, this conclusion is no longer true if the system is not fully isolated and interacts with its environment. The coupling with the environment can cause transitions ˆ c , as we shall see in the next section. between diﬀerent eigenstates of H Section 17.4: A Quantum Thermometer 17.4.1. (a) The sudden jumps of the signal are associated with a change of the cyclotron excitation number due to the coupling of the trapped electron with the thermostat. We recall that otherwise nr would be a constant of the motion. (b) In the case of the experimental curve a of Fig. 1, the fractions of time spent in the nl = 0, 1 and 2 levels are 80%, 19% and 1%, respectively. In the case of curve b, the fractions are 97% and 3% for the time spent on nl = 0 and nl = 1, respectively. 17.4.2. For a one-dimensional harmonic oscillator of frequency ω, one ﬁnds: pn+1 e−(n+3/2)¯hω/kB T = −(n+1/2)¯hω/k T = e−¯hω/kB T , B pn e which is independent of n. 17.4.3. For the curves of Fig. 1, we ﬁnd: ¯ ωc / |ln(0.24)| 0.7 ¯hωc . This corre• Curve a: p1 /p0 = 0.24, i.e. kB Ta = h sponds to Ta 5 K. In principle, the determination of the temperature cans also be made using p2 /p1 , but the accuracy is poor compared to that obtained with p1 /p0 . ¯ ωc / |ln(0.03)| 0.29 h ¯ ωc . This corre• Curve b: p1 /p0 = 0.03, i.e. kB Tb = h sponds to Tb 2 K. 17.4.4. (a) The normalization factor is determined from: 1=N

∞

e−(n+1/2)¯hω/kB T .

n=0

This is a geometric series in e−γ : 1=N

e−γ/2 1 − e−γ

→

N = 2 sinh(γ/2) .

182

17 A Quantum Thermometer

The mean excitation number is: ne−nγ d −nγ n ln nPn = −nγ = − e n ¯= dγ ne n n or: n ¯=

1 . eγ − 1

(b) One can see on the above expression that n ¯ is a rapidly increasing function of the temperature. If the temperature is such that γ ∼ 1, i.e. kB T ∼ ¯hωc (or T ∼ 7.1 K for this experiment), the mean excitation number is of the order of (e − 1)−1 ∼ 0.6. Below this temperature, the occupation of the level nl = 0 becomes predominant, as can be seen on the curves of Fig. 17.1. The variation of ln pn as a function of n is linear: ln pn = −n

hωc ¯ + Const kB T

The slope increases as the temperature decreases. The curves of Fig. 2 clearly show this linear variation. They correspond to ratios p1 /p0 equal 0.16 , 0.092 , 0.028 , 0.012 , i.e. temperatures of 1.6 K, 2 K, 3 K and 3.9 K, respectively. (c) In order to measure a temperature with such a device, one must use a statistical sample which is signiﬁcantly populated in the level nl = 1. It is experimentally diﬃcult to go below a probability of 10−2 for the level nr = 1, which corresponds to a temperature T ∼ 1.5 K. (d) In order to improve the sensitivity of this thermometer, one can: • Increase signiﬁcantly the total time of measurement in order to detect occupation probabilities of the level nr = 1 signiﬁcantly less than 10−2 ; • Reduce the value of the magnetic ﬁeld B, in order to reduce the cyclotron frequency ωc , and to increase (for a given temperature) the occupation probability of the level nl = 1 .

The data used in this chapter are extracted from the article of S. Peil and G. Gabrielse, Observing the Quantum Limit of an Electron Cyclotron: QND Measurements of Quantum Jumps between Fock States, Physical Review Letters 83, p. 1287 (1999).

Part III

Complex Systems

19 Properties of a Bose–Einstein Condensate

By cooling down a collection of integer spin atoms to a temperature of less than one micro-Kelvin, one can observe the phenomenon of Bose–Einstein condensation. This results in a situation where a large fraction of the atoms are in the same quantum state. Consequently, the system possesses remarkable coherence properties. We study here the ground state of such an N particle system, hereafter called a condensate. We will show that the nature of the system depends crucially on whether the two-body interactions between the atoms are attractive or repulsive.

19.1 Particle in a Harmonic Trap We consider a particle of mass m placed in a harmonic potential with a frequency ω/2π. The Hamiltonian of the system is ˆ2 1 ˆ = p + mω 2 rˆ 2 , H 2m 2 ˆ = (ˆ where rˆ = (ˆ x, yˆ, zˆ) and p px , pˆy , pˆz ) are respectively the position and momentum operators of the particle. We set a0 = ¯h/(mω). 19.1.1. Recall the energy levels of this system, and its ground state wave function φ0 (r). 19.1.2. We wish to obtain an upper bound on this ground state energy by the variational method. We use a Gaussian trial wave function: 1 ψσ (r) = exp(−r2 /(2σ 2 )) with σ > 0 . (19.1) 3/4 2 (σ π) The values of a relevant set of useful integrals are given below. By varying σ, ﬁnd an upper bound on the ground state energy. Compare the bound with the exact value, and comment on the result.

194

19 Properties of a Bose–Einstein Condensate

Formulas:

|ψσ (r)|2 dx dy dz = 1

|ψσ (r)|4 dx dy dz =

∂ψσ (r) 2 1 ∂x dx dy dz = 2σ 2

σ2 x |ψσ (r)| dx dy dz = 2 2

1 1 (2π)3/2 σ 3

2

19.2 Interactions Between Two Conﬁned Particles We now consider two particles of equal masses m, both placed in the same harmonic potential. We denote the position and momentum operators of the ˆ1, p ˆ2. two particles by rˆ 1 , rˆ 2 and p 19.2.1. In the absence of interactions between the particles, the Hamiltonian of the system is ˆ 21 ˆ2 p 1 1 ˆ = p H + 2 + mω 2 rˆ 21 + mω 2 rˆ 22 . 2m 2m 2 2 (a) What are the energy levels of this Hamiltonian? (b) What is the ground state wave function Φ0 (r 1 , r 2 )? 19.2.2. We now suppose that the two particles interact via a potential v(r 1 − r 2 ). We assume that, on the scale of a0 , this potential is of very short range and that it is peaked around the origin. Therefore, for two functions f (r) and g(r) which vary appreciably only over domains larger than a0 , one has

4π¯ h2 a f (r 1 ) g(r 2 ) v(r 1 − r 2 ) d r1 d r2 m 3

3

f (r) g(r) d3 r .

(19.2)

The quantity a, which is called the scattering length, is a characteristic of the atomic species under consideration. It can be positive (repulsive interaction) or negative (attractive interaction). One can measure for instance that for sodium atoms (isotope 23 Na) a = 3.4 nm, whereas a = −1.5 nm for lithium atoms (isotope 7 Li). (a) Using perturbation theory, calculate to ﬁrst order in a the shift of the ˆ caused by the interaction between the two ground state energy of H atoms. Comment on the sign of this energy shift. (b) Under what condition on a and a0 is this perturbative approach expected to hold?

19.4 Condensates with Repulsive Interactions

195

19.3 Energy of a Bose–Einstein Condensate We now consider N particles conﬁned in the same harmonic trap of angular frequency ω. The particles have pairwise interactions through the potential v(r) deﬁned by (19.2). The Hamiltonian of the system is N N N 2 ˆ p 1 1 i ˆ = H + mω 2 rˆ 2i v(ˆ r i − rˆ j ) . + 2m 2 2 i=1 j=1 i=1 j=i

In order to ﬁnd an (upper) estimate of the ground state energy of the system, we use the variational method with factorized trial wave functions of the type: Ψσ (r 1 , r 2 , . . . , r N ) = ψσ (r 1 ) ψσ (r 2 ) . . . ψσ (r N ) , where ψσ (r) is deﬁned in (19.1). 19.3.1. Calculate the expectation values of the kinetic energy, of the potential energy and of the interaction energy, if the N particle system is in the state |Ψσ : Ek (σ) = Ψσ |

N pˆi 2 i=1

2m

|Ψσ

Ep (σ) = Ψσ |

i=1 N

Eint (σ) = Ψσ |

N 1

2

mω 2 rˆi 2 |Ψσ

N

1 v(ˆ r i − rˆ j )|Ψσ 2 i=1 j=1 j=i

ˆ σ . We set E(σ) = Ψσ |H|Ψ ˜ 19.3.2. We introduce the dimensionless quantities E(σ) = E(σ)/(N ¯hω) and ˜ in terms of σ ˜ . Cast the result in the form σ ˜ = σ/a0 . Give the expression of E

3 1 η 2 ˜ E(σ) = + σ ˜ + 3 4 σ ˜2 σ ˜ and express the quantity η as a function of N , a and a0 . In all what follows, we shall assume that N 1. ˆ 19.3.3. For a = 0, recall the ground state energy of H.

19.4 Condensates with Repulsive Interactions In this part, we assume that the two-body interaction between the atoms is repulsive, i.e. a > 0.

196

19 Properties of a Bose–Einstein Condensate

˜ as a function of σ 19.4.1. Draw qualitatively the value of E ˜ . Discuss the ˜min . variation with η of the position of its minimum E 19.4.2. We consider the case η 1. Show that the contribution of the kinetic ˜ is negligible. In that approximation, calculate an approximate energy to E ˜min . value of E 19.4.3. In this variational calculation, how does the energy of the condensate vary with the number of atoms N ? Compare the prediction with the experimental result shown in Fig. 19.1. 19.4.4. Figure 19.1 has been obtained with a sodium condensate (mass m = 3.8 × 10−26 kg) in a harmonic trap of frequency ω/(2π) = 142 Hz. ¯ ω for this potential. (a) Calculate a0 and h (b) Above which value of N does the approximation η 1 hold? (c) Within the previous model, calculate the value of the sodium atom scattering length that can be inferred from the data of Fig. 19.1. Compare the result with the value obtained in scattering experiments a = 3.4 nm. Is it possible a priori to improve the accuracy of the variational method?

Fig. 19.1. Energy per atom E/N in a sodium condensate, as a function of the number of atoms N in the condensate

19.5 Condensates with Attractive Interactions We now suppose that the scattering length a is negative. ˜ as a function of σ 19.5.1. Draw qualitatively E ˜. 19.5.2. Comment on the approximation (19.2) in the region σ → 0.

19.6 Solutions

197

˜ no 19.5.3. Show that there exists a critical value ηc of |η| above which E longer has a local minimum for a value σ ˜ = 0. Calculate the corresponding size σc as a function of a0 . 19.5.4. In an experiment performed with lithium atoms (m = 1.17 × 10−26 kg), it has been noticed that the number of atoms in the condensate never exceeds 1200 for a trap of frequency ω/(2π) = 145 Hz. How can this result be explained?

19.6 Solutions Section 19.1: Particle in a Harmonic Trap 19.1.1. The Hamiltonian of a three-dimensional harmonic oscillator can be written ˆ =H ˆx + H ˆy + H ˆz , H ˆ i represents a one dimensional harmonic oscillator of same frequency where H ˆ of along the axis i = x, y, z. We therefore use a basis of eigenfunctions of H the form φ(x, y, z) = χnx (x) χny (y) χnz (z), i.e. products of eigenfunctions of ˆy, H ˆ z , where χn (x) is the nth Hermite function. The eigenvalues of H ˆ ˆ x, H H hω, where n = nx + ny + nz is a non-negative can be written as En = (n + 3/2)¯ integer. The ground state wave function, of energy (3/2)¯ hω, corresponds to nx = ny = nz = 0, i.e. 1 φ0 (r) = 2 3/4 exp[−r2 /(2a20 )] . (a0 π) 19.1.2. The trial wave functions ψσ are normalized. In order to obtain an ˆ we must calculate E(σ) = upper bound for the ground-state energy of H, ˆ σ and minimize the result with respect to σ. Using the formulas ψσ |H|ψ given in the text, one obtains pˆ2 1 h2 1 ¯ mω 2 σ 2 2 2 |ψσ = 3 mω | r |ψ ψ =3 ψσ | σ σ 2 2m 2m 2σ 2 2 2 and E(σ) =

3 hω ¯ 4

a20 σ2 + 2 2 σ a0

.

¯ ω. In This quantity is minimum for σ = a0 , and we ﬁnd Emin (σ) = (3/2) h this particular case, the upper bound coincides with the exact result. This is due to the fact that the set of trial wave functions contains the ground state ˆ wave function of H.

198

19 Properties of a Bose–Einstein Condensate

Section 19.2: Interactions between Two Conﬁned Particles ˆ 2 , where H ˆ1 ˆ can be written as H ˆ =H ˆ1 + H 19.2.1. (a) The Hamiltonian H ˆ and H2 are respectively the Hamiltonians of particle 1 and particle 2. A basis ˆ is formed by considering products of eigenfunctions of eigenfunctions of H ˆ 2 (functions of ˆ of H1 (functions of the variable r 1 ) and eigenfunctions of H hω, where n is a the variable r 2 ). The energy eigenvalues are En = (n + 3)¯ non-negative integer. ˆ is: (b) The ground state of H Φ0 (r 1 , r 2 ) = φ0 (r 1 ) φ0 (r 2 ) =

1 a30 π 3/2

" # exp −(r12 + r22 )/(2a20 ) .

ˆ is non-degenerate, its shift to ﬁrst 19.2.2. (a) Since the ground state of H order in a can be written as ∆E = Φ0 |˜ v |Φ0 = |Φ0 (r 1 , r 2 )|2 v(r 1 − r 2 ) d3 r1 d3 r2 1 1 4π¯ h2 a 4π¯ h2 a |φ0 (r)|4 d3 r = m m (2π)3/2 a30 therefore ∆E = hω ¯

2 a . π a0

For a repulsive interaction (a > 0), there is an increase in the energy of the system. Conversely, in the case of an attractive interaction (a < 0), the ground state energy is lowered. (b) The perturbative approach yields a good approximation provided the ˆ Therefore, energy shift ∆E is small compared to the level spacing ¯hω of H. one must have |a| a0 , i.e. the scattering length must be small compared to the spreading of the ground state wave function. Section 19.3: Energy of a Bose–Einstein Condensate 19.3.1. Using the formulas provided in the text, one obtains: Ek (σ) = N

¯2 3 h 4 mσ 2

Ep (σ) = N

3 mω 2 σ 2 4

aa2 2 N (N − 1) ¯hω 30 . Eint (σ) = 2 π σ Indeed, there are N kinetic energy and potential energy terms, and N (N −1)/2 pairs which contribute to the interaction energy.

19.6 Solutions

199

19.3.2. With the change of variables introduced in the text, one ﬁnds

3 1 N −1 a 1 2 ˜ E(σ) = +σ ˜ + √ 2 4 σ ˜ ˜3 2π a0 σ so that

N −1 a . η= √ 2π a0 19.3.3. If the scattering length is zero, there is no interaction between the particles. The ground state of the system is the product of the N functions φ0 (r i ) and the ground state energy is E = (3/2)N ¯hω.

Section 19.4: Condensates with Repulsive Interactions ˜ σ ) as a function of σ 19.4.1. Figure 19.2 gives the variation of E(˜ ˜ for increasing values of η. The value of the function for a given value of σ ˜ increases as η ˜ does not depend on η. It is dominated increases. For large σ ˜ , the behavior of E 2 by the potential energy term 3˜ σ /4. ˜min increases as η increases. This minimum corresponds The minimum E the point where the potential energy term, which tends to favor small values of σ, matches the kinetic and interaction energy terms which, on the contrary, favor large sizes σ. Since the interactions are repulsive, the size of the system is larger than in the absence of interactions, and the corresponding energy is also increased.

˜ σ ) with σ Fig. 19.2. Variation of E(˜ ˜ for η = 0, 10, 100, 1000 (from bottom to top)

19.4.2. Let us assume η is much larger than 1 and let us neglect a priori σ 2 + η/˜ σ 3 is minimum for the kinetic energy term 1/˜ σ 2 . The function (3/4)˜ σ ˜min = (2η)1/5 where its value is ˜min = 5 (2η)2/5 . E 4 One can check a posteriori that it is legitimate to neglect the kinetic energy term 1/˜ σ 2 . In fact it is always smaller than one of the two other contributions ˜ to E:

200

19 Properties of a Bose–Einstein Condensate

• For σ ˜σ ˜min , one has 1/˜ 19.4.3. For a number of atoms N 1, the energy of the system as calculated by the variational method is 5 E = ¯ hω N 4

a 2 N π a0

2/5 .

(19.3)

This variation of E/N as N 2/5 is very well reproduced by the data. In Fig. 19.3 we have plotted a ﬁt of the data with this law. One ﬁnds E/N αN 2/5 with α = 8.2 × 10−33 Joule.

Fig. 19.3. Fit of the experimental data with an N 2/5 law

19.4.4. (a) One ﬁnds a0 = 1.76 µm and h ¯ ω = 9.4 10−32 Joule. (b) Consider the value a = 3.4 nm given in the text. The approximation η 1 will hold if N 1300. This is clearly the case for the data of Fig. 19.1. (c) The coeﬃcient α = 8.2 × 10−33 Joule found by ﬁtting the data leads to a = 2.8 nm. This value is somewhat lower than the expected value a = 3.4 nm. This is due to the fact that the result (19.3), E/(N ¯hω) 1.142 (N a/a0 )2/5 , obtained in a variational calculation using simple Gaussian trial functions, does not yield a suﬃciently accurate value of the ground state energy. With more appropriate trial wave functions, one can obtain, in the mean ﬁeld aphω) 1.055 (N a/a0 )2/5 . The ﬁt proximation and in the limit η 1: Egs /(N ¯ to the data is then in agreement with the experimental value of the scattering length.

19.6 Solutions

201

Section 19.5: Condensates with Attractive Interactions ˜ σ ) is represented in Fig. 19.4. We notice that it has 19.5.1. The function E(˜ a local minimum only for small enough values of η. For η < 0, there is always a minimum at 0, where the function tends to −∞. 10

E (σ ) 5 0 -5 -10

σ 0

1

2

3

˜ σ ) for η = 0; η = −0.1; η = −0.27; η = −1 (from top to Fig. 19.4. Plot of E(˜ bottom)

19.5.2. The absolute minimum at σ = 0 corresponds to a highly compressed atomic cloud. For such small sizes, approximation (19.2) for a “short range” potential loses its meaning. Physically, one must take into account the formation of molecules and/or atomic aggregates which have not been considered here. ˜ σ ) has an inﬂexion 19.5.3. The local minimum at σ ˜ = 0 disappears when E(˜ point where the derivative vanishes. This happens for a critical value of η determined by the two conditions: ˜ dE =0 d˜ σ

˜ d2 E =0. d˜ σ2

This leads to the system 2η 1 +1− 5 4 σ ˜ σ ˜ 3 8η 0 = 4 +1+ 5 σ ˜ σ ˜ 0=−

from which we obtain |ηc | =

2 55/4

0.27

σ ˜c =

1 51/4

0.67

202

19 Properties of a Bose–Einstein Condensate

or σc 0.67 a0 . If the local minimum exists, i.e. for |η| < |ηc |, one can hope to obtain a metastable condensate, whose size will be of the order of the minimum found in this variational approach. On the other hand, if one starts with a value of |η| which is too large, for instance by trying to gather too many atoms, the condensate will collapse, and molecules will form. 19.5.4. For the given experimental data one ﬁnds a0 = 3.1 µm, and a critical number of atoms: √ a0 ∼ 1400 , Nc = 2π ηc |a| in good agreement with experimental observations.

19.7 Comments The ﬁrst Bose–Einstein condensate of a dilute atomic gas was observed in Boulder (USA) in 1995 (M.H. Anderson, J.R. Ensher, M.R. Matthews, C.E. Wieman, and E.A. Cornell, Science 269, 198 (1995)) with rubidium atoms. The experimental data shown in this chapter for a sodium condensate come from the results published by M.-O. Mewes, M.R. Andrews, N.J. van Druten, D.M. Kurn, D.S. Durfee, and W. Ketterle, Phys. Rev. Lett. 77, 416 (1996). The measurement of the energy E/N is done by suddenly switching oﬀ the conﬁning potential and by measuring the resulting ballistic expansion. The motion of the atoms in this expansion essentially originates from the conversion of the potential energy of the atoms in the trap into kinetic energy. The experimental results on lithium have been reported by C. Bradley, C.A. Sackett, and R.G. Hulet, Phys. Rev. Lett. 78, 985 (1997). The Nobel prize 2001 has been awarded to E. Cornell, W. Ketterle, and C. Wieman for the achievement of Bose-Einstein condensation in dilute gases of alkali atoms, and for early fundamental studies of the properties of the condensates.

20 Magnetic Excitons

Quantum ﬁeld theory deals with systems possessing a large number of degrees of freedom. This chapter presents a simple model, where we study the magnetic excitations of a long chain of coupled spins. We show that one can associate the excited states of the system with quasi-particles that propagate along the chain. We recall that, for any integer k: N

e2iπkn/N = N if k = pN, with p integer;

n=1

= 0 otherwise.

20.1 The Molecule CsFeBr3 Consider a system with angular momentum equal to 1, i.e. j = 1 in the basis |j, m common to Jˆ2 and Jˆz . 20.1.1. What are the eigenvalues of Jˆ2 and Jˆz ? 20.1.2. For simplicity, we shall write |j, m = |σ, where σ = m = 1, 0, −1. Write the action of the operators Jˆ± = Jˆx ± iJˆy on the states |σ. 20.1.3. In the molecule Cs Fe Br3 , the ion Fe2+ has an intrinsic angular momentum, or spin, equal to 1. We write the corresponding observable Jˆ , and we note |σ the eigenstates of Jˆz . The molecule has a plane of symmetry, and the magnetic interaction Hamiltonian of the ion Fe2+ with the rest of the molecule is ˆ r = D Jˆ2 H D>0. z h2 ¯ ˆ r and the corresponding energy values? Are What are the eigenstates of H there degeneracies?

204

20 Magnetic Excitons

20.2 Spin–Spin Interactions in a Chain of Molecules We consider a one-dimensional closed chain made up with an even number N of Cs Fe Br3 molecules. We are only interested in the magnetic energy states of the chain, due to the magnetic interactions of the N Fe2+ ions, each with spin 1. We take {|σ1 , σ2 , · · · , σN }, σn = 1, 0, −1, to be the orthonormal basis of ˆn the states of the system; it is an eigenbasis of the operators {Jˆzn } where J is the spin operator of the n-th ion (n = 1, · · · , N ). ˆ = The magnetic Hamiltonian of the system is the sum of two terms H ˆ ˆ H0 + H1 where N ˆ0 = D H (Jˆzn )2 h2 n=1 ¯ ˆ 1 is a nearest-neighbor spin–spin interaction has been introduced in 1.3, and H term N ˆ n+1 ˆn · J ˆ1 = A J A>0. H h2 n=1 ¯ ˆ N +1 ≡ J ˆ 1. ˆ 1 , we deﬁne J To simplify the notation of H ˆ 0 (A D), ˆ 1 is a small perturbation compared to H We assume that H and we shall treat it in ﬁrst order perturbation theory. ˆ 0 , and give the cor20.2.1. Show that |σ1 , σ2 , · · · , σN is an eigenstate of H responding energy value. ˆ 0 ? Is it a degenerate level? 20.2.2. What is the ground state of H ˆ 0 ? What is the 20.2.3. What is the energy of the ﬁrst excited state of H 1 degeneracy d of this level? We shall denote by E the corresponding eigenspace ˆ 0 , of dimension d. of H ˆ 1 can be written as 20.2.4. Show that H N

1 ˆn ˆn+1 ˆ1 = A ˆn Jˆn+1 ) + Jˆn Jˆn+1 . J ( J H + J + − − + z z h2 n=1 2 ¯

20.3 Energy Levels of the Chain We now work in the subspace E 1 . We introduce the following notation |n, ± = |σ1 = 0, σ2 = 0, · · · , σn = ±1, σn+1 = 0, · · · , σN = 0. Owing to the periodicity of the chain, we deﬁne |N + 1, ± ≡ |1, ±.

20.3 Energy Levels of the Chain

205

20.3.1. Show that ˆ 1 |n, ± = A(|n − 1, ± + |n + 1, ±) + |ψn , H where |ψn is orthogonal to the subspace E 1 . Without giving the complete form of |ψn , give an example of one of its ˆ 0 to which |ψn belongs. components, and give the energy of the eigenspace of H 20.3.2. Consider the circular permutation operator Tˆ, and its adjoint Tˆ† , deﬁned by Tˆ|σ1 , σ2 , · · · , σN = |σN , σ1 , · · · , σN −1 Tˆ† |σ1 , σ2 , · · · , σN = |σ2 , σ3 , · · · , σN , σ1 . Write the action of Tˆ and Tˆ† on the states |n, ±. ˆ 1 and A(Tˆ + Tˆ† ) have the same 20.3.3. Check that, in the subspace E 1 , H matrix elements. 20.3.4. Show that the eigenvalues λk of Tˆ are the N -th roots of unity (we recall that N is assumed to be even): λk = e−iqk

qk = −π +

2kπ N

k = 0, · · · , N − 1 .

20.3.5. We seek, in E 1 , the 2N eigenvectors |qk , ± of Tˆ, each corresponding to an eigenvalue λk . Each |qk , ± is written cn (k) |n, ± . (20.1) |qk , ± = n

(a) Write a recursion relation between the coeﬃcients cn . (b) Show that 1 cn (k) = √ eiqk n N is a solution of this recursion relation.

(20.2)

(c) Show that the states |qk , ± deﬁned using (20.1) and (20.2) are orthonormal. (d) Show that the vectors |qk , ± are also eigenvectors of Tˆ† and Tˆ + Tˆ† , and give the corresponding eigenvalues. (e) Calculate the scalar product n, |qk , (, = ±) and write the expansion of the states |n, ± in the basis |qk , ±. ˆ 1 of Sect. 2 as a perturbation to H ˆ 0. 20.3.6. We treat the Hamiltonian H ˆ We limit ourselves to the ﬁrst excited level of H0 , and we want to calculate how the perturbation lifts the degeneracy of this level. We recall that, in the degenerate case, ﬁrst order perturbation theory consists in diagonalizing the restriction of the perturbing Hamiltonian in the degenerate subspace ˆ 0. of the dominant term H

206

20 Magnetic Excitons

(a) Explain why the results of questions 3.3 and 3.5 above allow one to solve this problem. (b) In ﬁrst order perturbation theory, give the new energy levels which arise ˆ 0 , and the corresponding eigenstates. from the ﬁrst excited state of H (c) Draw qualitatively the energies E(qk ) in terms of the variable qk which can be treated as a continuous variable, qk ∈ [−π, +π[, if N is very large. 1. What is the degeneracy of each new energy level?

20.4 Vibrations of the Chain: Excitons We now study the time evolution of the spin chain. 20.4.1. Suppose that at time t = 0, the system is in the state |Ψ (0) =

−1 N =± k=0

ϕk |qk ,

with

−1 N

|ϕk |2 = 1 .

=± k=0

Setting ω = 2A/¯ h, write the state |Ψ (t) at a later time t. 20.4.2. We assume that the initial state is |Ψ (0) = |qk , +. (a) Write the probability amplitude αn (t) and the probability Pn (t) of ﬁnding at time t the n-th spin pointing upwards, i.e. σn = +1 and σm = 0 for m = n. Show that Pn (t) is the same for all sites of the chain. (b) The molecules of the chain are located at xn = na, where a is the lattice spacing. Show that the probability amplitude αn (t) is equal to the value at x = xn of a monochromatic plane wave Ψk (x, t) = Cei(p(q)x−E(q)t)/¯h , where C is a constant, q = qk , and x is the abscissa along the chain. Express p(q) in terms of q. (c) Show that Ψk (x, t) is an eigenstate of the momentum operator pˆx = (¯ h/i)∂/∂x along the chain. Show that the value of p(q) ensures the periodicity of Ψk (x, t), i.e. Ψk (x + L, t) = Ψk (x, t), where L = N a is the length of the chain. (d) Show that, for |qk | 1, Ψk (x, t) satisﬁes a Schr¨odinger equation for a particle of negative mass m, placed in a constant potential; give the value of m. 20.4.3. In a more complete analysis, one can associate quasi-particles to the magnetic excitations of the chain. These quasi-particles, which we call “magnetic excitons”, have an energy E(qk ) and a momentum p(qk ). At very low temperatures, T ≈ 1.4 K, the chain is in the ground state ˆ 0 . If low energy neutrons collide with it, they can create excitons whose of H energy and momentum can be determined by measuring the recoil of the neutrons. The experimental result for E(q) as a function of q ∈ [−π, 0] is given in Fig. 20.1.

20.4 Vibrations of the Chain: Excitons

207

Fig. 20.1. Experimental measurement of the excitation energy E(q) as a function of q between −π and 0. The energy scale is in meV (10−3 eV)

(a) Deduce from that data approximate values for D and A. (b) What do you think of the approximation D A and of the comparison between theory and experiment? How could one improve the agreement between theory and experiment? (c) Is it justiﬁed to assume that the chain is in its ground state when it is at thermal equilibrium at 1.4 K? We recall the Boltzmann factor: N (E2 )/N (E1 ) = exp[−(E2 − E1 )/kT ], with k = 8.6 × 10−5 eV K−1 . 20.4.4. Consider, at time t = 0, the state |Ψ (0) =

N −1 k=0

ϕk |qk , +

with

N −1

|ϕk |2 = 1.

k=0

We assume that N 1, that the coeﬃcients ϕk have signiﬁcant values only in a close vicinity of some value k = k0 , or, equivalently, q ≈ q0 , and that, to a good approximation, in this vicinity, dE . E(q) = E(q0 ) + (q − q0 )u0 , u0 = dq q=q0 Show that the probability Pn (t) of ﬁnding σn = +1 at time t is the same as the probability Pn (t ) of ﬁnding σn = +1 at another time t whose value will be expressed in terms of t and of the distance between the sites n and n . Interpret the result as the propagation of a spin excitation wave along the chain. Calculate the propagation velocity of this wave and give its numerical value for a = 0.7 nm and q0 = −π/2. 20.4.5. We now assume that the initial state is |Ψ (0) = |n = 1, +.

208

20 Magnetic Excitons

(a) Write the probability Pm (t) of ﬁnding σm = +1 at a later time t? (b) Calculate the probabilities P1 (t) and P2 (t), in the case N = 2, and interpret the result. (c) Calculate P1 (t) in the case N = 8. Is the evolution of P1 (t) periodic? (d) For N 1, one can convert the above sums into integrals. The prob2 where the Jn (x) are the Bessel abilities are then Pm (t) ≈ |Jm−1 (ωt)| 2 (x)| = 1 and Jn = (−)n J−n . functions. These functions satisfy |J n

2 For x 1 we have Jn (x) ≈ πx cos(x − nπ/2 − π/4) if x > 2|n|/π, and Jn (x) ≈ 0 if x < 2|n|/π. Which sites are appreciably reached by the probability wave at a time t such that ωt 1? (e) Interpret the result as the propagation along the chain of a probability amplitude (or wave). Calculate the propagation velocity and compare it with the result obtained in question 4.4).

20.5 Solutions Section 20.1: The Molecule CsFeBr3 20.1.1. The results are: Jˆ2 : 2¯ h2 , Jˆz : m¯h ; m = 1, 0, −1. 20.1.2. One has: J+ |1 = 0√ J+ |0 =h ¯ √2|1 J+ | − 1 = h ¯ 2|0

√ =h ¯ √2|0 J− |1 J− |0 =h ¯ 2| − 1 J− | − 1 = 0 .

20.1.3. The eigenstates are the states |σ. The state |0 corresponds to the eigenvalue E = 0, whereas |+ and |−, which are degenerate, correspond to E = D. Section 20.2: Spin–Spin Interactions in a Chain of Molecules 20.2.1. It is straightforward to see that ˆ 0 |σ1 , σ2 · · · σN = D H

N n=1

the corresponding eigenvalue being E = D

(σn )2 |σ1 · · · σN ,

σn2 .

ˆ 0 corresponds to all the σn equal to zero, so 20.2.2. The ground state of H that E = 0. This ground state is non-degenerate. 20.2.3. The ﬁrst excited state corresponds to all the σ’s being zero except one: σn equal to ±1. The energy is D, and the degeneracy 2N , since there are N possible choices of the non-vanishing σn , and two values ±1 of σn . 20.2.4. J± = Jx ± iJy . A direct calculation leads to the result.

20.5 Solutions

209

Section 20.3: Energy Levels of the Chain 20.3.1. The action of the perturbing Hamiltonian on the basis states is, setting = ±: ˆ 1 |n, = A (|n − 1, + |n + 1, ) H +A (|0, · · · 0, σn = , 0 · · · 0, σn = −1, σn +1 = +1, 0 · · · 0 n =n

+|0 · · · 0, σn = , 0 · · · 0, σn = +1, σn +1 = −1, 0 · · · 0) . The vector |ψ = |σ1 = 1, σ2 = −1, 0 · · · 0, σn = , 0 · · · 0 belongs to this latter ˆ 0 with energy 3D. set; it is an eigenvector of H 20.3.2. The deﬁnition of Tˆ, Tˆ† and |n, ± implies: Tˆ|n, ± = |n + 1, ± ;

Tˆ† |n, ± = |n − 1, ± .

20.3.3. We therefore obtain A(Tˆ + Tˆ† )|n, ± = A(|n − 1, ± + |n + 1, ±) . Since ˆ 1 |n, ± = A(|n − 1, ± + |n + 1, ±) + |ψn H

where n , ±|ψn = 0 ,

ˆ 1 and A(Tˆ + Tˆ† ) obviously have the same matrix elements in the subspace H E 1. ˆ an eigenvalue λk satisﬁes λN = 1, which proves that 20.3.4. Since TˆN = I, k each eigenvalue is an N -th root of unity. Conversely, we will see in the following that each N -th root of unity is an eigenvalue. 20.3.5. (a) The corresponding eigenvectors satisfy |qk , ± =

cn |n, ±

Tˆ|qk , ± = λk |qk , ±

n

therefore one has

cn |n + 1, ± = λk

n

cn |n, ± .

n

Hence the recursion relation and its solution are λk cn = cn−1

cn =

1

c1 λn−1 k

= eiqk (n−1) c1 .

210

20 Magnetic Excitons

(b) The √ normalization condition n |cn |2 = 1 gives N |c1 |2 = 1. If we choose c1 = eiqk / N , the eigenvectors are of unit norm and we recover the solution given in the text of the problem. (c) The scalar product of |qk , and |qk , is easily calculated: qk , |qk , = δ,

1 2iπn(k−k )/N e = δ, δk,k . N n

(d) The vectors |qk , ± are eigenvectors of Tˆ† with the complex conjugate eigenvalues λ∗k . Therefore they are also eigenvectors of Tˆ + Tˆ† with the eigenvalue λk + λ∗k = 2 cos qk = −2 cos(2kπ/N ). (e) From the deﬁnition of the vectors, we have 1 n, |qk , = √ eiqk n δ N and (directly or by using the closure relation) N −1 1 −iqk n e |qk , ± . |n, ± = √ N k=0

ˆ 1 to the subspace E 1 is identical to A(Tˆ + Tˆ† ) 20.3.6. The restriction of H 1 (question 3.3). In E , the operator A(Tˆ + Tˆ† ) is diagonal in the basis |qk , ±. ˆ 1 is also diagonal in that basis. The energy levels Therefore the restriction of H are (20.3) E(qk ) = D + 2A cos(qk ) , corresponding to the states |qk , ±. As far as degeneracies are concerned, there is a twofold degeneracy for all levels (the spin value may be +1 or −1). In addition, for all levels except q = −π and q = 0, there is a degeneracy qk ↔ −qk (symmetry of the cosine). Therefore, in general, the degeneracy is 4. Section 20.4: Vibrations of the Chain: Excitons 20.4.1. At time t the state of the chain is (cf. (20.3)): |Ψ (t) = e−iDt/¯h ϕk e−iωt cos qk |qk , .

k

20.4.2. We now consider an initial state |qk , ±, evolving as e−iE(q)t/¯h |qk , ±. (a) We therefore obtain an amplitude 1 αn (t) = √ ei(qk n−E(qk )t/¯h) N and a probability Pn (t) = |αn |2 =

1 N,

which is the same on each site.

20.5 Solutions

211

(b) In the expression 1 αn (t) = √ ei(qk xn /a−E(qk )t/¯h) , N we see that αn (t) is the value at x = xn of the function Ψk (x, t) = √1 N

exp[i(px − Et)/¯ h] with E(q) = D + h ¯ ω cos q and p(q) = h ¯ q/a.

(c) The function Ψk (x) is an eigenstate of pˆx with the eigenvalue ¯hqk /a. Since N is even, we obtain: eiqk L/a = eiN qk = e2πik = 1 , which proves the periodicity of Ψk . (d) For |qk | 1, cos qk = 1 − qk2 /2 . Therefore E = E0 + p2 /2m with E0 = D + 2A

and

m=−

¯2 h ¯h =− 2 . 2Aa2 ωa

Ψk then satisﬁes the wave equation i¯ h

h2 ∂ 2 ψ ¯ ∂ψ =− + E0 ψ , ∂t 2m ∂x2

which is a Schr¨ odinger equation for a particle of negative mass (in solid state physics, this corresponds to the propagation of holes and in ﬁeld theory, to the propagation of anti-particles). 20.4.3. (a) With the data of the ﬁgure which resemble grosso modo the E(q) drawn in Fig. 20.2, one ﬁnds D + 2A ∼ 3.2 × 10−3 eV, and D − 2A ∼ 0.4 × 10−3 eV. Therefore: D ∼ 1.8 × 10−3 eV

A ∼ 0.7 × 10−3 eV .

(b) The approximation D A is poor. The theory is only meaningful to order (A/D)2 ∼ 10%. Second order perturbation theory is certainly necessary to account quantitatively for the experimental curve which has a steeper shape than a sinusoid in the vicinity of q = −π. (c) For T = 1.4 K, kT ∼ 1.2 × 10−4 eV, exp(−(D − 2A)/kT ) ∼ 0.04. To a few % , the system is in its ground state. 20.4.4. Approximating E(q) = E(q0 ) + (q − q0 )u0 in the vicinity of q0 , we obtain 1 ϕk ei(qk −q0 )(n−u0 t/¯h) . αn (t) = √ ei(q0 n−ω0 t) N k

212

20 Magnetic Excitons

Fig. 20.2. Energy levels

Since the global phase factor does not contribute to the probability, one has Pn (t) = Pn (t ) with h ¯ t = t + (n − n) . u0 This corresponds to the propagation of a wave along the chain, with a group velocity a dE u0 a 2a A = sin q0 . =− vg = h ¯ h dq q=q0 ¯ ¯h For q0 = −π/2 and a = 0.7 nm, we ﬁnd vg ∼ 1500 ms−1 . One can also evaluate u0 ∼ 1.2 meV directly on the experimental curve, which leads to vg ∼ 1300 ms−1 . √ −iqk 20.4.5. If |Ψ (0) = |n = 1, +, then ϕ+ / N and ϕ− k =e k = 0. (a) The probability is Pm (t) = |m, +|Ψ (t)|2 , where m, +|Ψ (t) =

e−iDt/¯h iqk (m−1) −iωt cos qk e e . N k

(b) N=2: There are two possible values for qk : q0 = −π and q1 = 0. This leads to P1 (t) = cos2 ωt, P2 (t) = sin2 ωt. These are the usual oscillations of a twostate system, such as the inversion of the ammonia molecule. (c) N=8: qk cos(qk )

−π

− 3π 4

−1

− √12

− π2

− π4

0

0

√1 2

1

π 4 √1 2

π 2

0

3π 4 − √12

The probability P1 of ﬁnding the excitation on the initial site is P1 (t) =

√ 2 1 2 cos (ωt/2) + cos(ωt/ 2) . 4

The system is no longer periodic in time. There cannot exist t = 0 for which

20.5 Solutions

213

√ P1 (t) = 1, otherwise there would exist n and n such that 2 = n /n. (d) Since Jn (ωt) ∼ 0 for ωt < 2|n|/π, only sites for which |m − 1| < πωt/2 are reached at time t. For large ωt, the amplitude is the same for all sites of the same parity: Pm (t) =

2 π π cos2 ωt − (m − 1) − . πωt 2 4

We notice in particular that Pm (t) + Pm+1 (t) = 2/(πωt) is independent of m and varies slowly with t. (e) The probability wave becomes delocalized very quickly on the chain (ωt > a few π). The edges of the region where the probability is non zero propagate in opposite directions with the velocity v = πωa/2. This is comparable with what we have found in 4.4 for a wave packet near q = π/2. The experimental data displayed in this chapter were obtained by B. Dorner et al., Z. Phys. B 72, 487 (1988).

21 A Quantum Box

In recent years, it has become possible to devise quantum boxes (also called quantum dots) of nanometric dimensions, inside which the conduction electrons of a solid are conﬁned at low temperatures. The ensuing possibility to control the energy levels of such devices leads to very interesting applications in micro-electronics and opto-electronics. A quantum box is made of a material A on which another material B is deposited. A set of quantum boxes is shown on Fig. 21.1. The dots of In As (material B) are deposited on a substate of Ga As (material A).

Fig. 21.1. Picture of a set of quantum boxes obtained with a tunneling microscope. The side of the square is 1 µm long and the vertical scale will be studied below

216

21 A Quantum Box

In this chapter, we are interested in the motion of an electron in a twodimensional box. We note −q the electric charge of the electron, and we neglect spin eﬀects. We shall assume that in a solid, the dynamics of an electron is described by the usual Schr¨ odinger equation where: (i) the mass of the electron is replaced by an eﬀective mass µ, (ii) the atoms of the materials A and B create an eﬀective potential V (x, y) which is slowly varying on the atomic scale.

21.1 Results on the One-Dimensional Harmonic Oscillator Consider a particle of mass µ placed in the one-dimensional potential V (x) = µω 2 x2 /2. We recall the deﬁnition of the annihilation and creation operators ˆ†x of the oscillator in terms of the position and momentum operators a ˆx and a x ˆ and pˆx

pˆx pˆx µω µω 1 1 † + i√ − i√ a ˆx = √ a ˆx = √ . (21.1) x ˆ x ˆ h ¯ ¯h hµω ¯ ¯hµω 2 2 The Hamiltonian of the system can be written as

2 1 ˆ x = pˆx + 1 µω 2 x H ˆ2 = h ¯ω n ˆx + where 2µ 2 2

ˆ†x a ˆx . n ˆx = a

(21.2)

We also recall that the eigenvalues of the number operator n ˆ x are the nonnegative integers. Noting |nx the eigenvector corresponding to the eigenvalue nx , we have √ √ a ˆx |nx = nx |nx − 1 (21.3) a ˆ†x |nx = nx + 1|nx + 1 21.1.1. We recall that the ground state wave function is

µω 1/4 µωx2 ψ0 (x) = exp − . π¯ h 2¯h What is the characteristic extension 0 of the electron’s position distribution in this state? 21.1.2. The eﬀective mass µ of the electron in the quantum box is µ = 0.07 m0 , where m0 is the electron mass in vacuum. We assume that ¯hω = 0.060 eV, i.e. ω/(2π) = 1.45 × 1013 Hz. (a) Evaluate 0 numerically. (b) At a temperature of 10 Kelvin, how many levels of the oscillator are populated signiﬁcantly? (c) What is the absorption wavelength of radiation in a transition between two consecutive levels?

21.2 The Quantum Box

217

21.2 The Quantum Box We assume that the eﬀective two-dimensional potential seen by an electron in the quantum box is: V (x, y) =

1 2 2 µω x + y 2 . 2

(21.4)

ˆ 0 = (ˆ We note H p2x + pˆ2y )/2µ + V (x, y) the Hamiltonian of the electron. ˆ†y and n ˆ y in an analogous way to (21.1) 21.2.1. We deﬁne the operators a ˆy , a and (21.2). Give a justiﬁcation for the fact that the states |nx , ny , which ˆ y with integer eigenvalues nx and ny , form an are eigenstates of n ˆ x and n ˆ 0 in terms of nx and ny . ˆ 0 . Give the energy levels EN of H eigenbasis of H 21.2.2. What is the degeneracy gN of each level EN where N = 0, 1, 2 . . . ? ˆz = x 21.2.3. Express the operator L ˆpˆy − yˆpˆx as a function of the operators † † ˆx , a ˆy , a ˆy . a ˆx , a ˆ 0 . Do the states ˆ z on the eigenstates |nx , ny of H 21.2.4. Write the action of L |nx , ny have a well-deﬁned angular momentum Lz ? ˆ 0. 21.2.5. We are now interested in ﬁnding another eigenbasis of H ˆ ˆ (a) Show that H0 and Lz commute. Interpret this result physically. (b) We introduce the “left” and “right” annihilation operators as 1 ax + iˆ a ˆl = √ (ˆ ay ) 2

1 ax − iˆ a ˆr = √ (ˆ ay ) 2

(21.5)

ˆ†r . Write the commutation and the corresponding creation operators a ˆ†l , a relations of these four operators. ˆ†l a ˆl and n ˆr = a ˆ†r a ˆr commute. Using the values of the (c) Show that n ˆl = a † commutators [ˆ al , a ˆl ] and [ˆ ar , a ˆ†r ], and following the same procedure as in the usual quantization of the harmonic oscillator, justify that the eigenˆ l and n ˆ r are integers. values nl and nr of n ˆ r } form a complete set of commuting observables in We assume that {ˆ nl , n the problem under consideration, and we note {|nl , nr } the corresponding eigenbasis. ˆ r in terms of a ˆx , a ˆ†x , a ˆy and a ˆ†y . Deduce (d) Write the expression of n ˆ l and n ˆ z in terms of n ˆ 0 and L from that an expression of the operators H ˆ l and n ˆr . ˆ 0 and Justify that the states {|nl , nr } form an eigenbasis common to H ˆz. L ˆ z and H ˆ 0 . What are the allowed (e) We note m¯h and EN the eigenvalues of L values for the quantum number m in a given energy level EN ? (f) Represent the allowed couples of quantum numbers (m, N ) by points in the (Lz , EN ) plane. Show that one recovers the degeneracy of the levels ˆ 0. of H

218

21 A Quantum Box

ˆ 0 generated by |nx = 1, ny = 0 21.2.6. We consider the eigen-subspace of H and |nx = 0, ny = 1 as deﬁned in Sect. 2.1 above. Write the eigenstates of ˆ z in this basis and give the corresponding eigenvalues. L

21.3 Quantum Box in a Magnetic Field One applies a uniform magnetic ﬁeld B, parallel to the z axis, on the quantum box. This ﬁeld derives from the vector potential A(r) = −(yB/2)ux + (xB/2)uy where ux and uy are the unit vectors along the x and y directions. We assume that the Hamiltonian of the electron in the quantum box and in presence of the ﬁeld is given by 1 1 ˆ B = 1 (ˆ H px + qAx (ˆ (ˆ py + qAy (ˆ r ))2 + r ))2 + µω 2 (ˆ x2 + yˆ2 ) 2µ 2µ 2 with the usual canonical commutation relations [ˆ x, pˆx ] = [ˆ y , pˆy ] = i¯h and y , pˆx ] = 0. We introduce the cyclotron angular frequency ωc = qB/µ, [ˆ x, pˆy ] = [ˆ (ωc > 0). ˆ B and show that one can always ﬁnd a basis of the sys21.3.1. Expand H tem for which both the total energy and the angular momentum along z are simultaneously well deﬁned. 21.3.2. We deﬁne Ω = ω 2 + ωc2 /4. By redeﬁning in a simple manner the operators of Sect. 2, give the energy levels Enl ,nr of the system in terms of the two integers nl and nr . 21.3.3. For simplicity we subtract the zero point energy ¯hΩ from the energies ˜n ,n = En ,n − ¯ hΩ. Enl ,nr , and we set E r r l l ˜n ,n in the two limits of (a) Give the approximate expressions of the levels E r l a weak and of a strong ﬁeld, and give a deﬁnition of these regimes. (b) Plot as a function of the magnetic ﬁeld, the positions of the energy levels originating from the N = 0, 1, 2 levels in the absence of magnetic √ ﬁeld. (c) Show that two levels cross for a value of B such that ωc = ω/ 2 and specify which states they correspond to. √ 21.3.4. In the following we assume that ωc < ω/ 2. Using the values of ω and of the eﬀective mass µ given in the ﬁrst section of the problem, determine which values of the magnetic ﬁeld correspond to this inequality. ˆ B have respective energies 21.3.5. Show that the ﬁrst three eigenstates of H ¯Ω , E0 = h

E− = 2¯ hΩ −

hωc ¯ , 2

E+ = 2¯ hΩ +

¯hωc . 2

We note |u0 , |u− , |u+ the three corresponding eigenstates. What is the value of the angular momentum Lz in each of these three states?

21.4 Experimental Veriﬁcation

219

21.4 Experimental Veriﬁcation One can study the energy levels of an electron in a quantum box by measuring the absorption spectrum of a light beam. Absorption peaks appear at the Bohr frequencies (Ef −Ei )/h of the box, corresponding to excitations of the electron from an initial level |ui to a ﬁnal one |uf . 21.4.1. At a temperature of 10 K, one observes that only the level |ui = |u0 contributes signiﬁcantly to the absorption signal. Justify this fact using the result of question 1.2. 21.4.2. The experimental values of the frequencies of the ﬁrst two absorption peaks of a quantum box are given in Fig. 21.2, according to the values of the applied magnetic ﬁeld. Verify that the model developed above accounts for the slope of the curves for suﬃciently large values of B, but that it fails to describe the behavior near B = 0.

absorption frequency (1012 Hz)

17 16 15 14 13 12 0

2

4

6

8

10

Magnetic field (Tesla) Fig. 21.2. Frequencies ν = ω/2π of the two ﬁrst absorption peaks of a quantum box as a function of the magnetic ﬁeld B

21.4.3. Role of the z Dimension The conﬁnement along the z direction can be simulated by an inﬁnite onedimensional square well potential of extension D in this direction. (a) Recall the energy levels of a particle of mass µ in a one-dimensional inﬁnite well of width D. What is the energy diﬀerence between the two lowest lying states?

220

21 A Quantum Box

(b) Under what condition between D and ω is it legitimate to consider that the motion along the z axis is “frozen”, and to restrict to the lowest lying levels of the harmonic motion in the x, y plane? (c) On the picture 21.1, the vertical (z) and horizontal (xy) scales are diﬀerent. In order to neglect the motion along the z direction, which of the two scales must be contracted? One can make use of the value of l0 calculated in Sect. 1.2.

21.5 Anisotropy of a Quantum Box In order to reproduce the positions of the absorption peaks for low values of the ﬁeld (see Fig. 21.2), we assume that the conﬁning potential in the quantum box is slightly anisotropic. We therefore replace the expression (21.4) of the conﬁning potential by V (x, y) =

1 1 2 µω (1 + )x2 + µω 2 (1 − )y 2 2 2

with 1 .

(21.6)

We treat the problem using perturbation theory. In what follows, we asˆ B, can sume that the magnetic ﬁeld is small (ωc ω). The Hamiltonian H therefore be written as: ˆ B, = H ˆ0 + W ˆ H with 2 ˆ2 1 ˆ z + µω (ˆ ˆ0 = p ˆ = ωc L H + µω 2 (ˆ x2 − yˆ2 ) . x2 + yˆ2 ) and W 2µ 2 2 2

ˆ is the perturbation, and we neglect terms of order B 2 . In this expression, W We work in the basis {|nx , ny }. 21.5.1. Using the potential given in (21.6), determine the energy levels of the electron for B = 0. 21.5.2. Applying perturbation theory, evaluate to ﬁrst order in B and the energy shift of the ground state of the quantum box with respect to the value found in Sect. 2. 21.5.3. We are now interested in the position of the two energy states origiˆ 0 found in Sect. 2. nating from the ﬁrst excited level of H (a) Write the restriction of the Hamiltonian HB, in the basis {|nx = 1, ny = 0 , |nx = 0, ny = 1} of the corresponding subspace. (b) Deduce the approximate values of the energies of interest E− (B, ) and E+ (B, ) (with E− < E+ ). (c) Give the expressions of the corresponding eigenstates |u− (B, ) and |u+ (B, ). One can introduce the mixing angle α such that tan(2α) = ωc /(ω).

21.6 Solutions

221

21.5.4. We now turn back to the experimental data of Fig. 21.2. (a) Is the transition region between large values of B and B = 0 correctly described by this model of an anisotropic quantum box? (b) What value of the anisotropy can be deduced from the positions of the ﬁrst two absorption peaks in the absence of a magnetic ﬁeld?

21.6 Solutions Section 21.1: Results on the One-Dimensional Harmonic Oscillator 21.1.1. The characteristic spatial√extension of the position distribution is h/(µω). More precisely, 0 / 2 is the r.m.s. deviation of the probability 0 = ¯ law |ψ0 (x)|2 for the position distribution. 21.1.2. (a) One ﬁnds 0 = 4.3 nm. (b) The ratio of populations in the ﬁrst excited level n = 1 and the ground state n = 0 is given by the Boltzmann law r = exp(−¯ hω/(kB T )) = exp(−70) = 5 × 10−31 , which is negligible. The population of other excited levels is even smaller. Therefore, at T = 10 K, only the ground state is populated. (c) One ﬁnds λ = 2πc/ω = 21 µm, which corresponds to an infrared radiation. Section 21.2: The Quantum Box ˆ0 = h 21.2.1. The Hamiltonian is H ¯ ω(ˆ nx + n ˆ y + 1), and the operators n ˆ x and n ˆ y commute. We can ﬁnd an eigenbasis common to the two operators. If a x dependence is completely defunction Ψ (x, y) is an eigenfunction of n ˆ x , its termined (Hermite function of the variable x µω/¯h, corresponding to |nx ). ˆ y } is therefore a CSCO, with Similarly, for the y dependence. The set {ˆ nx , n ˆ 0 , the eigenvalue coreigenbasis |nx , ny . This basis is also an eigenbasis of H ¯ ω(N + 1) with N = nx + ny . responding to |nx , ny is EN = h 21.2.2. The energy level EN = h ¯ ω(N + 1) corresponds to N + 1 possible couples for (nx , ny ): (N, 0), (N − 1, 1), . . ., (0, N ). The degeneracy is therefore gN = N + 1.

222

21 A Quantum Box

21.2.3. We replace x ˆ, pˆx , yˆ, pˆy by their values:

h † ¯ hµω † ¯ a ˆx + a a ˆx − a pˆx = i ˆx ˆx x ˆ= 2µω 2

...

which leads to:

†

†

†

h † ˆ z = i¯ L a ˆx + a ˆy − a ˆy + a ˆx − a ˆx a ˆx a ˆy − a ˆy a ˆx = i¯h a ˆ†y − a ˆ†x a ˆy . 2 21.2.4. Making use of the action of the creation and annihilation operators, one ﬁnds: ˆ z |nx , ny = i¯ L h nx (ny + 1) |nx − 1, ny + 1 − (nx + 1)ny |nx + 1, ny − 1 . ˆ z and does not have a A state |nx , ny is not, in general, an eigenstate of L well deﬁned angular momentum. The only exception is the state for which ˆ z |0, 0 = 0. It is a state with Lz = 0. nx = ny = 0, and L ˆ 0. 21.2.5. We are now interested in ﬁnding another eigenbasis of H ˆ ˆ (a) The commutator of H0 and Lz can be calculated using their expressions in terms of the creation and annihilation operators. We ﬁrst calculate ˆ†x ] = a ˆ†x and [ˆ nx , a ˆx ] = −ˆ ax , and the analogous relations for y. One [ˆ nx , a obtains: ˆ 0, L ˆ z ] = i¯ [H h2 ω [ˆ nx + n ˆy , a ˆx a ˆ†y − a ˆ†x a ˆy ]

nx , a ˆx ]a†y − [ˆ nx , a ˆ†x ]ˆ ay + a ˆx [ˆ ny , a†y ] − a ˆ†x [ˆ ny , a ˆy ] = i¯ h2 ω [ˆ

ax a†y − a ˆ†x a ˆy + a ˆx a†y + a ˆ†x a ˆy = i¯ h2 ω −ˆ =0. This result is the consequence of the rotation invariance around the z axis of the potential V (x, y). This can also be proven by using the polar coordinate ∂ ˆ z = −i¯ ˆ0 expression: L h ∂φ , which commutes with the two contributions to H (kinetic and potential energy). (b) There are 4 operators, and therefore 6 commutators to evaluate. We ﬁrst remark that the two creation operators commute, and so do the two annihilation operators. Therefore: ˆ†r ] = 0 , [ˆ a†l , a A simple calculation leads to: 1 a ˆl , a ax + iˆ ˆ†l = [ˆ ay , a ˆ†x − iˆ a†y ] = 2 # 1 " ax − iˆ ˆ†r = [ˆ ay , a ˆ†x + iˆ a†y ] = a ˆr , a 2

[ˆ al , a ˆr ] = 0 .

1 [ˆ ax , a ˆ†x ] + 2 1 [ˆ ax , a ˆ†x ] + 2

1 [ˆ ay , a ˆ†y ] = 1 , 2 1 [ˆ ay , a ˆ†y ] = 1 . 2

21.6 Solutions

223

Finally, one ﬁnds: # 1 " ax + iˆ ˆ†r = [ˆ ay , a ˆ†x + iˆ a†y ] = a ˆl , a 2 1 a ˆr , a ax − iˆ ˆ†l = [ˆ ay , a ˆ†x − iˆ a†y ] = 2

1 [ˆ ax , a ˆ†x ] − 2 1 [ˆ ax , a ˆ†x ] − 2

1 [ˆ ay , a ˆ†y ] = 0 , 2 1 [ˆ ay , a ˆ†y ] = 0 . 2

Therefore, any “right” and “left” operators commute. The commutation relations between right (or left) creation and annihilation operators are the same as for usual one-dimensional creation and annihilation operators. ˆ r is obvious, (c) The commutation between the hermitian operators n ˆ l and n since any left operator commutes with any right operator. The quantization of the one-dimensional harmonic oscillator is entirely based on the commutation ˆ† a ˆ are relation [ˆ a, a ˆ† ] = 1, which leads to the fact that the eigenvalues of a ˆr , the non-negative integers. The same holds here for the operators n ˆ l and n the corresponding eigenvalues are the couples of integers (nl , nr ). As it is suggested in the text, we assume for the moment that the common eigenbasis ˆ r is unique, and we note |nl , nr the eigenvector corresponding to of n ˆ l and n the couple of eigenvalues (nl , nr ). (d) One ﬁnds:

1 † 1 (ˆ ax − iˆ n ˆx + n a†y ) (ˆ ax + iˆ ay ) = ˆ y + i(ˆ a†x a ˆy − a ˆx a ˆ†y ) 2 2 ˆ ˆ Lz 1 H0 −1− , = 2 ¯ hω h ¯

1 † 1 ax + iˆ n ˆx + n n ˆ r = (ˆ a†y ) (ˆ ax − iˆ ay ) = ˆ y − i(ˆ a†x a ˆy − a ˆx a ˆ†y ) 2 2 ˆ0 ˆz L 1 H −1+ , = 2 ¯ hω h ¯ n ˆl =

therefore ˆ0 = h ˆz = h H ¯ ω (ˆ nl + n ˆ r + 1) , L ¯ (ˆ nr − n ˆl) . ˆ 0 and L ˆ z can be expressed in terms of the sole operators The operators H ˆ r . The eigenbasis common to n ˆ l and n ˆ r is therefore also a common n ˆ l and n ˆz. ˆ 0 and L eigenbasis of H ˆ 0 and L ˆ z , with eigenvalues E = (e) The vector |nl , nr is the eigenstate of H ¯ (nr − nl ). We therefore have N = nl + nr and hω(nl + nr + 1) and Lz = h ¯ m = nr − nl . We recover the integer values for the orbital angular momentum, as expected. For a given N , the value of m belongs to the set {−N, −N + 2, . . . , N − 2, N }, therefore there are N + 1 possible values. We remark that in an energy level EN , m has the same parity as N . This comes from the parity invariance of the problem under consideration. (f ) The graphical representation is given on Fig. 21.3. On a given horizontal line corresponding to a given energy value, one ﬁnds N + 1 points, which

224

21 A Quantum Box

E / hω 4 3 2 1 Lz / h -3

-2

-1

0

1

2

3

Fig. 21.3. Allowed quantum numbers for the couple Lz , E

ˆ 0 found previously. This correspond to the degeneracy of an energy level of H ˆ r } form a CSCO. If two diﬀerent states would justiﬁes the fact that {ˆ nl , n correspond to the same couple of eigenvalues (nl , nr ), the corresponding point of the diagram would be twofold degenerate, and the degeneracy of the energy level EN would be larger than N + 1. ˆ z corresponding to 21.2.6. We must ﬁnd in this subspace two eigenvectors of L the two eigenvalues ±¯ h. A ﬁrst method for ﬁnding these eigenvectors consists ˆ z on the vectors of the basis {|nx , ny }. Using in calculating the action of L ˆ z in terms of a ˆx , a ˆy , . . ., one ﬁnds: the expression of L ˆ z |nx = 1, ny = 0 = i¯ L h |nx = 0, ny = 1 ˆ z |nx = 0, ny = 1 = −i¯ L h |nx = 1, ny = 0 ,

0 −i¯ h or the 2 × 2 matrix to diagonalize: . i¯ h 0 The eigenstates associated to the eigenvalues ±¯ h are therefore: √ (|nx = 1, ny = 0 ± i|nx = 0, ny = 1) / 2 . Another method consists in starting from the ground state |nl = 0, nr = 0 and letting act on this state: (i) the operator a ˆ†r in order to obtain the eigenvector of energy 2¯hω and angular momentum +¯ h, (ii) the operator a ˆ†l in order to obtain the eigenvector of energy 2¯ hω and angular momentum −¯ h. Of course, we recover the previous result.

21.6 Solutions

225

Section 21.3: Quantum Box in a Magnetic Field ˆ B , one ﬁnds: 21.3.1. By expanding H ˆB = H

pˆ2x + pˆ2y + 2µ

µω 2 q2 B 2 + 2 8

(ˆ x2 + yˆ2 ) +

ˆz ωc L . 2

ˆ z /2, ˆB = H ˆ (Ω) + ωc L 21.3.2. If we set Ω = ω 2 + ωc2 /4, we can rewrite H 0 (Ω) ˆ where H0 is the Hamiltonian of a two-dimensional oscillator of frequency Ω: pˆ2 + pˆ2y µΩ 2 2 ˆ (Ω) = x + (ˆ x + yˆ2 ) . H 0 2µ 2 One can then repeat the method of the previous section, by replacing ω by Ω in ˆy ,... One constructs an eigenbasis common the deﬁnition of the operators a ˆx , a (Ω) ˆ z , which we continue to note {|nl , nr }, the eigenvalues being to H0 and L ˆB, hΩ(nl + nr + 1) and m¯h. Each vector |nl , nr is also an eigenvector of H ¯ corresponding to the energy Enl ,nr = h ¯ Ω(nl + nr + 1) + h ¯ ωc (nr − nl )/2 ωc ωc ¯ Ω− ¯Ω . =h ¯ Ω+ nr + h nl + h 2 2 21.3.3. (a) Two limiting regimes of the magnetic ﬁeld can be considered, corresponding to the limits ωc ω (very weak magnetic ﬁeld) and ωc ω (very strong magnetic ﬁeld). In the ﬁrst case, we have in ﬁrst order in B: ˜n ,n ¯ E hω(nl + nr ) + h ¯ ωc (nr − nl )/2 , r l which corresponds to a linear variation in B of the N + 1 levels arising from the level EN in the absence of the ﬁeld. The slope (¯hqB/(2µ) (nr − nl ) is diﬀerent for each level, which means that there is no degeneracy if B does not vanish. For strong ﬁelds, one ﬁnds Ω+

ωc ωc , 2

Ω−

ω2 ωc ω . 2 ωc

We therefore have ˜n ,n ¯ E hωc nr r l

if nr = 0 ,

and

¯ ω2 ˜n ,0 h E nl . l ωc

For a non-vanishing nr , the energy level increases linearly with B, the slope ˜ tends to zero as 1/B. being proportional to nr . For nr = 0, the energy E ˜ (b) The energy levels Enl ,nr corresponding to N = 0, 1, 2 are represented on Fig. 21.4.

226

21 A Quantum Box

E/hΩ 2

1

0

ω /ω c 0

1

2

3

4

5

˜n ,nr arising from N = 0, 1, 2 as a function Fig. 21.4. Variation of the energy levels E l of the magnetic ﬁeld B

(c) We notice on Fig. 21.4 that the levels nl = 2, nr = 0 and nl = 0, nr = 1 cross each other. The corresponding value of the ﬁeld B is given by the solution of the equation ωc ωc =2 Ω− Ω+ , 2 2 √ or 3 ωc = 2 Ω, i.e. ωc = ω/ 2. √ 21.3.4. √ The value of the ﬁeld B which corresponds to ωc = ω/ 2 is µω/(q 2) 26 T. 21.3.5. If we assume that the ﬁeld B is smaller than 21 T, the three ﬁrst ˆ B correspond to nl = nr = 0 (ground state of energy energy levels of H hΩ − ¯ hωc /2), and nl = 0, nr = 1 (energy hΩ), nl = 1, nr = 0 (energy 2¯ ¯ ˆ z with the eigenvalues 2¯hΩ + h ¯ ωc /2). These three states are eigenstates of L 0, −¯ h and h ¯ respectively. Section 21.4: Experimental Veriﬁcation 21.4.1. In the absence of a magnetic ﬁeld, we saw in question 1.2 that only the level nx = ny = 0 is appreciably populated for T = 10 K. As the magnetic ﬁeld increases, the energy splitting between the ground state and the ﬁrst excited √ state diminishes√but it stays much smaller than kB T if ωc is less √ ¯ ω/ 2. For that value, the ratio than ω/ 2. For ωc = ω/ 2, the splitting is h between the populations of the ﬁrst excited state and the ground state is r = exp(−49) = 3.7 × 10−22 . Since only the ground state is populated, all the detectable absorption lines will occur from transition starting from this state. 21.4.2. The ﬁrst two absorption peaks correspond to the transitions |u0 ↔ |u− and |u0 ↔ |u+ . The corresponding frequencies ν± are such that ν± =

ωc Ω ± . 2π 4π

21.6 Solutions

227

The domain in B explored on the experimental ﬁgure of the text corresponds to values of ωc which are small compared to ω. We can therefore use the weak-ﬁeld expansion of question 3.3 in order to simplify this expression ωc ω ± . ν± = 2π 4π We therefore expect that the frequencies ν± will vary linearly with B, the slopes being ±q/(4πµ), and that the two straight lines will cross for a vanishing ﬁeld at the frequency ω/(2π). This linear variation of ν± does appear on the ﬁgure for higher values of B and the measured slope is close to the expected value (2 × 1011 Hz T−1 ). However, the experimentally observed behavior for a very weak ﬁeld does not correspond to our theoretical prediction. Instead of two lines of same frequency for B = 0, there is a ﬁnite diﬀerence ν+ − ν− . 21.4.3. Role of the z Dimension (a) The energy levels of an inﬁnite square well of size D are given by En = π 2 ¯h2 n2 /(2µD2 ), with n positive integer, the corresponding eigenstates are the functions χn (z) ∝ sin(nπz/D). The splitting between the ground state and h2 /(2µD2 ). the ﬁrst excited state is ∆E = 3π 2 ¯ (b) In order to consider that the motion along z is “frozen”, the energy splitting ∆E between the two ﬁrst levels of the square well must be very large compared to ¯hω. If this condition is satisﬁed, the accessible states for the electron conﬁned in the quantum box (in a reasonable domain of temperatures and exciting frequencies) will be simply combinations of the vectors |nx , ny ⊗ |χ0 . It is then legitimate to neglect the dynamics of the electron along z. If this condition is not satisﬁed, absorption lines can appear for frequencies near those presented on the experimental ﬁgure. They will correspond to the excitation of the motion along the z axis. The condition that the z motion be “frozen” is h2 3π 2 ¯ ¯ hω , or equivalently D π0 . (21.7) 2µD2 (c) In order for the harmonic approximation of the transverse motion to be valid, the transverse extension ∆L of the quantum box must be large compared to 0 . The condition obtained in the previous question D π0 , put together with 0 ∆L, imposes that the box must have a very ﬂat geometrical shape: the height D along z must be very small compared to its transverse extension in xy. We conclude that the vertical scale of the picture 21.1 is very dilated. Section 21.5: Anisotropy of a Quantum Box ˆ y with ˆx + H 21.5.1. For vanishing B, the Hamiltonian is H 2 ˆ x = pˆx + 1 µω 2 (1 + )ˆ H x2 , 2µ 2

ˆy = H

pˆ2y 1 + µω 2 (1 − )ˆ x2 . 2µ 2

228

21 A Quantum Box

ˆ x and H ˆ y , corresponding to products One can ﬁnd a common eigenbasis for H of Hermite functions in the variable x µω(1 + )/¯ h by Hermite functions in the variable y µω(1 − )/¯ h. The corresponding eigenvalues are √ √ ¯ hω (nx −ny ) hω 1 − (ny +1/2) ¯hω (nx +ny +1)+ hω 1 + (nx +1/2)+¯ ¯ 2 where nx , ny are non-negative integers. 21.5.2. To ﬁrst order in B and , the shift of the ground state energy is given by the matrix element 2 ˆ z |0, 0 + µω 0, 0|ˆ ˆ |0, 0 = ωc 0, 0|L x2 − yˆ2 |0, 0 . ∆E0,0 = 0, 0|W 2 2

ˆ z with eigenvalue 0. The ﬁrst term in this The state |0, 0 is an eigenstate of L y 2 |0, 0, sum therefore vanishes. By symmetry, we have 0, 0|ˆ x2 |0, 0 = 0, 0|ˆ which means that the second term also vanishes, to ﬁrst order in and B. ˆ z in the basis under 21.5.3. (a) We have already determined the matrix L consideration in question 2.6. We must calculate the matrix elements of x ˆ2 2 and yˆ . In order to do that, the simplest is to use the expressions of x ˆ and yˆ in terms of creation and annihilation operators. One has: x ˆ2 =

h ¯ (ˆ ax + a ˆ†x )(ˆ ax + a ˆ†x ) , 2µω

which leads to h ¯ 1, 0|ˆ a†x a ˆx + a ˆx a ˆ†x |1, 0 2µω 3¯h h ¯ (1 + 2) = , = 2µω 2µω 1, 0|ˆ x2 |0, 1 = 0, 1|ˆ x2 |1, 0 = 0 , h ¯ 0, 1|ˆ x2 |0, 1 = 0, 1|ˆ ax a ˆ†x |0, 1 2µω h ¯ , = 2µω

1, 0|ˆ x2 |1, 0 =

where we have set |0, 1 ≡ |nx = 0, ny = 1, etc. for simplicity. We obtain a ˆ B, in similar result by exchanging the roles of x and y. The restriction of H the subspace of interest is therefore

h ω −iωc ¯ ˆ hω + . [HB, ] = 2¯ 2 iωc − ω (b) The energy eigenvalues are obtained by diagonalizing this 2 × 2 matrix hω ± E± (B, ) = 2¯

¯ 2 2 h ω + ωc2 . 2

21.7 Comments

229

(c) Setting tan 2α = ωc /( ω), the above matrix is written as:

h 2 2 ¯ cos 2α −i sin 2α 2 ˆ hω + ω + ωc , [HB, ] = 2¯ i sin 2α − cos 2α 2 whose eigenvectors are

|u− (B, ) =

i sin α cos α

|u+ (B, ) =

cos α i sin α

.

21.5.4. (a) The variation of E± (B, ) − E0,0 with B reproduces well the experimental observations. For large values of B such that ω ωc , we recover the linear variation with B of the two transition frequencies. When B tends to zero (ωc ω), one ﬁnds two diﬀerent Bohr frequencies corresponding respectively to the two transitions nx = ny = 0 → nx = 0, ny = 1 and nx = ny = 0 → nx = 1, ny = 0. (b) When B tends to zero, one ﬁnds experimentally that the limit of (ν+ − ν− )/(ν+ + ν− ) is of the order of 0.06. The theoretical prediction for this ratio is /2. We therefore conclude that 0.12.

21.7 Comments Quantum boxes of semiconductors, a simple model of which has been examined here, are the subject of many investigations both academic (Coulomb correlations) and applied (optronics). Here we have only considered electronic excitations, but collective modes in a lattice (phonons) also play an important role in the dynamics of quantum boxes. It has been shown recently that the two types of excitations are strongly coupled. This is in contrast with the usual situation encountered in semiconductors, for which the coupling between the electrons and the phonons is weak. The data presented here come from S. Hameau et al., Phys. Rev. Lett. 83, 4152 (1999).

22 Colored Molecular Ions

Some pigments are made of linear molecular ions, along which electrons move freely. We derive here the energy levels of such an electronic system and we show how this energy scheme explains the observed color of the pigments. Consider molecular ions of the chemical formula (Cn Hn+2 )− , which can be considered as deriving from polyethylene molecules, such as hexatriene CH2 =CH-CH=CH-CH=CH2 , with an even number of carbon atoms, by removing a CH+ group. In an ion of this type, the bonds rearrange themselves and lead to a linear structure of the following type: −

(CH2 · · · CH · · · CH · · · CH · · · CH2 )

,

(22.1)

with an odd number n of equally spaced carbon atoms separated by d = 1.4 ˚ A. In this structure, one can consider that the n+1 electrons of the double bonds of the original polyethylene molecule move independently of one another in a one-dimensional inﬁnite potential well of length Ln = nd: V (x) = +∞ =0

for for

x < 0 or x > Ln 0 ≤ x ≤ Ln .

(22.2)

Actually, one should write Ln = (n − 1)d + 2b where b represents the edge eﬀects. Experimentally, the choice b = d/2 appears to be appropriate.

22.1 Hydrocarbon Ions 22.1.1. What are the energy levels εk of an electron in this potential? 22.1.2. Owing to the Pauli principle, at most two electrons can occupy the same energy level. What are the energies of the ground state E0 and of the ﬁrst excited state E 1nof the set of n + 1 electrons? We recall that k=1 k 2 = n(n + 1)(2n + 1)/6.

232

22 Colored Molecular Ions

22.1.3. What is the wavelength λn of the light absorbed in a transition between the ground state and the ﬁrst excited state? One can introduce the A. Compton wavelength of the electron: λC = h/(me c) = 2.426 × 10−2 ˚ 22.1.4. Experimentally, one observes that the ions n = 9, n = 11 and n = 13 A), yellow light (λ11 ∼ 6000 ˚ A) and red light absorb blue light (λ9 ∼ 4700 ˚ A), respectively. Is the previous model in agreement with this (λ13 ∼ 7300 ˚ observation? Are the ions n ≤ 7 or n ≥ 15 colored?

22.2 Nitrogenous Ions One can replace the central CH group by a nitrogen atom, in order to form ions of the type: −

(CH2 · · · CH · · · N · · · CH · · · CH2 )

.

(22.3)

The presence of the nitrogen atom does not change the distances between atoms but it changes the above square well potential. The modiﬁcation consists in adding a small perturbation δV (x), attractive and localized around the nitrogen atom: δV (x) = 0 = −V0

for for

|x − |x −

Ln 2 | Ln 2 |

> α/2 ≤ α/2 ,

where α/d 1 and V0 > 0. 22.2.1. Using ﬁrst order perturbation theory, give the variations δεk of the energy level εk of an electron in the well. For convenience, give the result to leading order in α/d. 22.2.2. Experimentally, one observes that, for the same value of n, the spectrum of the nitrogenous ions (22.3) is similar to that of the ions (22.1) but that the wavelengths λN n are systematically shorter (blue-shifted) if n = 4p+1, and systematically longer (red-shifted) if n = 4p + 3, than those λ0n of the corresponding hydrocarbons (22.1). Explain this phenomenon and show that λN n and λ0n are related by: n+1 λ0n n , = 1 − (−1) 2 γ N λn n+2

where γ is a parameter to be determined. ˚ 22.2.3. The nitrogenous ion n = 11 absorbs red light (λN 11 ∼ 6700 A). Check ˚ ∼ 4300 A ). What is the color of that the ion n = 9 absorbs violet light (λN 9 the nitrogenous ion n = 13? 22.2.4. For suﬃciently large n, if the nitrogen atom is placed not in the central site but on either of the two sites adjacent to the center of the chain,

22.3 Solutions

233

one observes the reverse eﬀect, as compared to question 2.2. There is a red shift for n = 4p + 1 and a blue shift for n = 4p + 3. Can you give a simple explanation for this eﬀect?

22.3 Solutions Section 22.1: Hydrocarbon Ions 22.1.1. The energy levels are εk =

h2 k 2 π2 ¯ 2mL2n

k = 1, 2, . . . .

22.1.2. The ground state energy of the n + 1 electrons is E0 =

h2 π2 ¯ mL2n

(n+1)/2

k2 =

k=1

h2 π2 ¯ (n + 1)(n + 2)(n + 3) . 24 mL2n

The energy of the ﬁrst excited state is E1 = E0 +

# h2 " π2 ¯ π 2 ¯h2 (n + 3)2 − (n + 1)2 = E0 + (n + 2) . 2 8 mLn 2 mL2n

22.1.3. One has hν = E1 − E0 = π 2 ¯ h2 (n + 2)/(2 mL2n ). Since λ = c/ν, we obtain an absorption wavelength λn =

8 d2 n2 . λC (n + 2)

22.1.4. From the general form λn = 646.33 n2 /(n + 2), we obtain λ9 = A, λ13 = 7280 ˚ A, in good agreement with experiment. 4760 ˚ A, λ11 = 6020 ˚ A and λ5 = 2310 ˚ A are in the For smaller n, the wavelengths λ7 = 3520 ˚ ultraviolet part of the spectrum. The ions n ≤ 7 do not absorb visible light and are thus not colored. A and λ17 = 9830 ˚ A are in the For n ≥ 15, the wavelengths λ15 = 8550 ˚ infrared region. These ions do not absorb visible light in transitions from the ground state to the ﬁrst excited state. They are nevertheless colored because of transitions to higher excited states. Section 22.2: Nitrogenous Ions 2/Ln sin(kπx/Ln ). 22.2.1. The normalized wave functions are ψk (x) = One has Ln +α/2 |ψk (x)|2 dx . δεk = δV (x) |ψk (x)|2 dx = −V0 Ln −α/2

234

22 Colored Molecular Ions

Setting y = x − Ln /2, one obtains δεk = −

2V0 Ln

+α/2

sin2 −α/2

kπ kπy + 2 nd

There are two cases: • k even:

2V0 +α/2 2 kπy sin δεk = − dy , Ln −α/2 nd

i.e.

dy .

3

δεk = O((α/d) ) .

The perturbation is negligible. • k odd:

kπy 2V0 +α/2 cos2 δεk = − dy . Ln −α/2 nd To ﬁrst order in α/d, we have δεk = −2V0 α/nd < 0. The exact formulas are: δεk = −

kπα V0 k Ln sin α − (−1) . Ln kπ Ln

The (single particle) energy levels corresponding to even values of k are practically unaﬀected by the perturbation; only those with k odd are shifted. This is simple to understand. For k even, the center of the chain is a node of the wave function, and the integral deﬁning δεk is negligible. For k odd, on the contrary, the center is an antinode, we integrate over a maximum of the wave function, and the perturbation is maximum. 22.2.2. The perturbation to the excitation energy E1 − E0 of question 1.2 is δE = δε(n+3)/2 − δε(n+1)/2 . • (n + 1)/2 even, i.e. n = 4p + 3, δε(n+1)/2 = 0, δE = δε(n+3)/2 = −

2V0 α 0. nd

We can summarize these results in the compact form

n+1 h2 n + 2 π2 ¯ n 2 E1 − E0 + δE = γ 1 − (−1) , 2md2 n2 n+2

22.4 Comments

235

with γ = 4V0 αmd/(π¯ h)2 . We therefore obtain the desired relation n+1 λ0n n . = 1 − (−1) 2 γ N λn n+2

For n = 4p+1, the perturbation increases the excitation energy, and decreases λn . For n = 4p + 3, it decreases the excitation energy, and increases λn . 22.2.3. For the ion n = 11 one obtains the relation (1 − 11γ/13) = 6000/6700, ˚ therefore γ ∼ 0.12 and λN 9 = 4330 A, in good agreement with experiment. One ˚ = 6600 A , which absorbs red light and gives a green color also obtains λN 13 to the corresponding pigment. Note that the presence of the nitrogen atom N 0 0 yields λN 13 ≤ λ11 whereas λ13 > λ11 . 22.2.4. The distance between a node and an antinode of ψk (x) is δx = nd/(2k). For k = (n + 1)/2 and k = (n + 3)/2 which are the states of interest, we will have respectively δx = nd/(n + 1) and δx = nd/(n + 3), i.e. δx ∼ d if n is large. Consequently, if a wave function has a node at the center, it has an antinode in the vicinity of the two adjacent sites, and vice versa. The argument is therefore similar to the answer to questions 2.1 and 2.2, with the reverse eﬀect. The lines are red-shifted if n = 4p + 1 and they are blue-shifted if n = 4p + 3.

22.4 Comments Many further details can be found in the article by John R. Platt, The Chemical Bound and the Distribution of Electrons in Molecules, D Conjugated Chains,, Handbuch Der Physik, Volume XXXVII/2, p. 173, Springer-Verlag (1961). This article is a very complete work on the applications of Quantum Mechanics in Chemistry.

23 Hyperﬁne Structure in Electron Spin Resonance

Many molecular species, such as free radicals, possess an unpaired electron. The magnetic spin resonance of this electron, called electron spin resonance (ESR) as opposed to nuclear magnetic resonance, provides useful information about the electronic structure of the molecule, as we shall see in this chapter. We assume here the following: 1. Spin variables and space variables are independent, both for electrons and for nuclei; we are only interested in the former. 2. The spatial ground state of the unpaired electron is non-degenerate, and one can neglect the eﬀect of a magnetic ﬁeld on its wave function. 3. We only take into account the following magnetic spin interactions: (a) the Zeeman interaction of spin magnetic moments with an external ﬁeld B, and (b) the hyperﬁne interaction between the outer electron and the nuclei. 4. For a given nucleus in the molecule, the hyperﬁne interaction has the form ˆ · Iˆ = (A/4)σ ˆ =h ˆ HF = (A/¯ ˆe · σ ˆ n where S ˆ e /2 is the electron spin h2 ) S ¯σ H ˆ ˆ e and σ ˆ n are the Pauli matrices ˆ n /2 is the nuclear spin; σ and I = h ¯σ which act respectively in the Hilbert spaces of the electron and of the nucleus. The constant A is given by 2 h2 |ψ(r n )|2 , A = − µ0 γe γn ¯ 3 where µ0 = 1/0 c2 is the magnetic susceptibility of vacuum, γe and γn are the gyromagnetic factors of the electron and of the nucleus under consideration, and ψ(r n ) is the value of the electron wave function at the position r n of this nucleus. 5. In all the problem, the system is considered to be in a constant uniform magnetic ﬁeld B directed along the z axis. For simplicity, we set A = h ¯ a, ωe = −γe B, ωn = −γn B and η = (ωe − ωn )/2. The numerical values of gyromagnetic ratios are

238

23 Hyperﬁne Structure in Electron Spin Resonance

electron: proton:

γe /(2π) = −28.024 GHz T−1 , γp /(2π) = +42.574 MHz T−1 .

23.1 Hyperﬁne Interaction with One Nucleus 23.1.1. We ﬁrst consider a species where the nuclei do not possess a magnetic moment, so that there is no hyperﬁne interaction. Write the Zeeman interaction Hamiltonian of the electron with the magnetic ﬁeld B. What are the energy levels of the system? What is the value of the frequency that can excite the system? Give its numerical value for a magnetic ﬁeld of 1 Tesla. 23.1.2. We now assume that the molecule has one spin-1/2 nucleus. We note the (factorized) eigenbasis common to Sˆz and Iˆz as {|σe ; σn } with σe = ±1 and σn = ±1. (a) Write the complete spin Hamiltonian. (b) Calculate the action of σ e · σ n on the vectors of the basis {|σe ; σn }. (c) Write the matrix form of the Hamiltonian in this basis, and calculate its eigenvalues. 23.1.3. From now on, we assume that the magnetic ﬁeld B is strong, in the sense that |ωe | |a|. (a) Give the approximate form of the eigenvalues to ﬁrst order in a/η. (b) Recover these results by ﬁrst diagonalizing the electron Zeeman Hamiltonian, and by treating the other terms, i.e. the nuclear Zeeman Hamiltonian and the hyperﬁne interaction, in ﬁrst order perturbation theory. What are the corresponding eigenstates (to zeroth order in a/η)? (c) One can show that the transitions that an electromagnetic ﬁeld can induce occur only between states which diﬀer by the value of a single spin (for instance, the transitions |+; − → |−; + are forbidden). Under these conditions, what are the observable transition frequencies, knowing that all transitions which are not forbidden actually occur? Classify these transitions in two sets corresponding respectively to nuclear and to electronic spin transitions. (d) Calculate these frequencies numerically for the hydrogen atom in a ﬁeld B = 1 T. We recall that, in this case, A/(2π¯h) 1.420 GHz.

23.2 Hyperﬁne Structure with Several Nuclei We now assume that the molecule has N protons in hydrogen atoms located on sites r 1 , . . . , r N , whose spins are denoted Iˆ 1 , . . . , Iˆ N . The Hilbert space of spin degrees of freedom is of dimension 2N +1 . It is spanned by the set:

23.3 Experimental Results

239

{|σe ; σ1 , σ2 , . . . , σN } ≡ {|σe ⊗ |σ1 ⊗ |σ2 ⊗ · · · ⊗ |σN } with σe = ±1 and σk = ±1, k = 1, . . . , N . This set is an orthonormal eigenbasis common to the z projection of the spin observables Sˆz and Iˆkz , k = 1, . . . , N , of the N + 1 particles. ¯ ak be the hyperﬁne constant of proton k. Write the ex23.2.1. Let Ak = h pression for the spin Hamiltonian of the system (we recall that the magnetic nucleus–nucleus interaction is neglected). 23.2.2. Show that the restriction of this Hamiltonian to each eigen-subspace of Sˆz is diagonal. 23.2.3. Assuming, as in 1.3, that the ﬁeld is strong, calculate the eigenvalues in ﬁrst order perturbation theory, and the corresponding eigenstates. 23.2.4. What are the observable electron spin transition frequencies? How many lines corresponding to these frequencies should the spectrum display in principle? 23.2.5. What is the number of lines and the multiplicity of each of them (i.e. the number of transitions at the same frequency) if all the protons are equivalent, i.e. if all the |ψ(r k )|2 , and therefore the coeﬃcients ak , are equal? 23.2.6. What is the number of lines and their multiplicities, if there exist two sets of equivalent protons, one with p protons corresponding to the constant ap , the other with q = N − p protons, corresponding to the constant aq ?

23.3 Experimental Results Experimentally, one measures the positions and the intensities of the absorption lines in the microwave region. An absorption line appears as a peak in the absorbed intensity α(ν) as a function of the frequency, whose qualitative shape is shown in Fig. 23.1. It can be shown that the intensity of an absorption peak at a given frequency is proportional to the number of transitions (multiplicity of the line) which can occur at that frequency. For experimental convenience, one ﬁxes the frequency of the microwave at a given value, and one varies the magnetic ﬁeld B. This results in an absorption curve α(B). 23.3.1. Figure 23.2 shows the spectrum of the free radical • CH3 (methyl) (J.N. Chazalviel, private communication). The carbon nucleus does not possess any magnetic moment; only the protons of the hydrogen atoms give rise to hyperﬁne interactions. (a) Interpret this spectrum qualitatively. Explain the number of lines and their relative intensities. How many diﬀerent coeﬃcients ak are there?

240

23 Hyperﬁne Structure in Electron Spin Resonance

Fig. 23.1. Typical shape of an ESR absorption curve as a function of the frequency

2,3 10-3 T

α (B) B Fig. 23.2. Microwave spectrum of the radical • CH3

(b) Give the value of ak /(2π). Calculate the value of |ψ(r k )|2 for the unpaired electron in this molecule. It is convenient to express the result in terms of |ψ(0)|2Hydrogen = 1/(πa31 ) where a1 is the Bohr radius of hydrogen. 23.3.2. Answer the same questions for the spectrum of CH3 −• COH−COO− (the radical ion of lactic acid) shown in Fig. 23.3. Neither the oxygen nor the carbon nuclei carry magnetic moments. The only hyperﬁne interaction arises, again, from the protons of the hydrogen atoms.

23.4 Solutions Section 23.1: Hyperﬁne Interaction with One Nucleus ˆ = −¯ ˆez /2, hence the energy 23.1.1. The magnetic Hamiltonian is H hγe B σ hγe B/2 corresponding to the states |± . The transition levels E± = ∓¯ ¯ ωe , ν = ωe /(2π). For B = 1 T, frequency is given by hν = E+ − E− = h ν = 28.024 GHz. 23.1.2. (a) The full Hamiltonian, including the hyperﬁne interaction, is

23.4 Solutions

241

1,7110-3 T αB

2 10-4 T

B

Fig. 23.3. Microwave spectrum of the radical CH3 −• COH−COO−

ˆ · Iˆ ˆ = −γe B Sˆz − γn B Iˆz + A S H h2 ¯ hωn ¯ ¯ha hωe ¯ ˆn . ˆe · σ σ ˆez + σ ˆnz + σ = 2 2 4 ˆ n on the basis states is: ˆe · σ (b) The action of σ ˆe · σ ˆ n |+; + = |+; + σ ˆe · σ ˆ n |+; − = 2|−; + − |+; − σ ˆe · σ ˆ n |−; + = 2|+; − − |−; + σ ˆ n |−; − = |−; − . ˆe · σ σ (c) Hence the 4 × 4 matrix representation of the Hamiltonian ⎛ ⎞ 0 0 a + 2(ωe + ωn ) 0 ⎟ h⎜ 0 4η − a 2a 0 ˆ =¯ ⎟ , ⎜ H ⎠ ⎝ 0 2a −4η − a 0 4 0 0 0 a − 2(ωe + ωn ) where the rows and columns are ordered as |+; +, |+; −, |−; +, |−; −. Hence the eigenstates and the corresponding eigenvalues: ¯ h (a + 2(ωe + ωn )) 4 h ¯ |−; − −→ (a − 2(ωe + ωn )) 4 |+; + −→

and from the diagonalization of the 2 × 2 matrix between |+; − and |−; + ¯ h (−a + 2 4η 2 + a2 ) 4 h ¯ sin φ |+; − − cos φ |−; + −→ (−a − 2 4η 2 + a2 ) 4

cos φ |+; − + sin φ |−; + −→

with tan φ =

a . 2η + 4η 2 + a2

242

23 Hyperﬁne Structure in Electron Spin Resonance

23.1.3. (a) If η a, the eigenvectors and eigenvalues are, to lowest order, |+; + |+; − |−; + |−; −

−→ (¯ h/4)(a + 2(ωe + ωn )) −→ ∼ (¯ h/4)(4η − a) −→ ∼ (¯ h/4)(−4η − a) −→ (¯ h/4)(a − 2(ωe + ωn )) .

(b) In each subspace corresponding respectively to σe = 1 and σe = −1, the perturbation is diagonal (the non-diagonal terms couple σe = +1 and σe = −1). The 2 × 2 matrices to be considered are indeed ˆ σn +, σn |H|+,

ˆ and −, σn |H|−, σn .

ˆ Consider for instance +, σn |H|+, σn . Since +, σn |Sˆx |+, σn = +, σn |Sˆy |+, σn = 0 , only +|Sˆz |+σn |Iˆz |σn has to be considered, and it is diagonal. The eigenstates at zeroth order are therefore |σe ; σn and we recover the above results. (c) Transitions: (i) Nuclear transitions: |σe ; + ↔ |σe ; − , i.e. |+; + ↔ |+; − |−; + ↔ |−; −

∆E = h ¯ (ωn + a/2), ∆E = h ¯ (ωn − a/2),

ν = |ωn + a/2|/(2π) ν = |ωn − a/2|/(2π) .

(ii) Electronic transitions: |+; σn ↔ |−; σn , i.e. |+; + ↔ |−; + |+; − ↔ |−; −

∆E = h ¯ (ωe + a/2), ∆E = h ¯ (ωe − a/2),

ν = |ωe + a/2|/(2π) ν = |ωe − a/2|/(2π) .

(d) For B = 1 T, νn = 42.6 MHz; a/(2π) = A/(2π¯h) = 1420 MHz; νe = 28.024 GHz. The nuclear transitions occur at ν1 = 753 MHz and ν2 = 667 MHz, the electronic transitions occur at ν1 = 28.734 GHz and ν2 = 27.314 GHz. Section 23.2: Hyperﬁne Structure with Several Nuclei 23.2.1. The total Hamiltonian is N

N

k=1

k=1

¯ Ak hωn hωe ˆ =¯ ˆe · σ ˆk . σ ˆez + σ ˆkz + σ H 2 2 4 ˆ to a subspace corresponding to the eigenvalue 23.2.2. The restriction of H ¯ σe /2 of Sˆez (σe = ±) can be written using 1.2(b) or (c): h N

hωn ¯ Ak σe hωe ˆσ = ¯ H σ + + σ ˆkz . e e 2 2 4 k=1

ˆ + and H ˆ − are diagonal in the basis {|σ1 , σ2 , . . . , σN }. The operators H

23.4 Solutions

243

23.2.3. First order Nperturbation theory N consists in diagonalizing the perturbˆ e ·σ ˆ k in each eigen-subspace hωn /2)ˆ σkz + k=1 (Ak /4)σ ing Hamiltonian k=1 (¯ ˆez /2. This is automatically satisﬁed. Therefore, of the dominant term h ¯ ωe σ σe = +1 : Eσ+1 ...σN =

h(2ωn + ak ) hωe ¯ ¯ + σk , state |+; σ1 , . . . , σN , 2 4 k

σe = −1 : Eσ−1 ...σN = −

h(2ωn − ak ) hωe ¯ ¯ + σk , state |−; σ1 , . . . , σN . 2 4 k

23.2.4. There are 2N transitions |+; σ1 , . . . , σN ↔ |−; σ1 , . . . , σN corresponding to the 2N possible choices for the set {σk }. The corresponding frequencies are 1 ∆νσ1 ...σN = ak σk /2 . ω e + 2π k

23.2.5. If all ak are equal to a, we have 1 1 |ωe + M a/2| , σk /2 = ∆ν = ω e + a 2π 2π k

with M = σk = N, N −2, . . . , −N +2, −N , i.e. N +1 absorption lines. There (N −M )/2 are CN transitions which have the same frequency and contribute to each line. The relative intensities of the lines will therefore be proportional to (N −M )/2 . The splitting between two adjacent lines the binomial coeﬃcients CN is a. 23.2.6. If p equivalent protons correspond to the coupling constant Ap , and q = N − p correspond to Aq , then p q 1 1 ap aq ap aq = ω ∆ν = + M + σ + σ + M ω . e i j e p q 2π 2 i=1 2 j=1 2π 2 2 There are p + 1 values of Mp : p, p − 2, . . . , −p, and q + 1 values of Mq : Mq = q, q −2, . . . , −q. The total number of lines is (p+1) (q +1), and the multiplicity (p−Mp )/2 (q−Mq )/2 Cq . of a line corresponding to a given couple (Mp , Mq ) is Cp Section 23.3: Experimental Results 23.3.1. The experimental results conﬁrm the above analysis. 1. For • CH3 there are 4 equally spaced lines of relative intensities 1 : 3 : 3 : 1. This is in perfect agreement with the fact that the three protons of • CH3 are obviously equivalent. All the Ak coeﬃcients are equal.

244

23 Hyperﬁne Structure in Electron Spin Resonance

2. For a ﬁxed ω, one gets by considering two consecutive lines, for instance the center lines: a/2 − γe B1 = −a/2 − γe B2 so that a = γe (B1 − B2 ). We h, and deduce ν = |a|/2π = 65 MHz = |Ak |/2π¯ πa31 |ψ(rk )|2 = |ψ(rk )|2 /|ψ(0)|2Hydrogen = 65/1420 ∼ 0.045 . In the radical • CH3 , the probability that the outer electron is on top of a proton is smaller by a factor 3 × 0.045 = 0.135 than in the hydrogen atom. 23.3.2. In the case of CH3 −• COH−COO− , there are four dominant lines, each of which is split into two. This agrees with the fact that, in the molecule CH3 −• COH−COO− , the 3 protons of the CH3 group are equivalent and have the same hyperﬁne constant a1 whereas the proton of the • COH group has a diﬀerent constant a2 which is noticeably smaller than a1 . A calculation similar to the previous one gives |ψ(rk )|2 /|ψ(0)|2Hydrogen ∼ 0.034 for the protons of the CH3 group, and |ψ(rk )|2 /|ψ(0)|2Hydrogen ∼ 0.004 for the proton of • COH.

24 Probing Matter with Positive Muons

A very eﬃcient technique for probing the structure of crystals consists in forming, inside the material, pseudo hydrogenic atoms made of an electron and a positive muon, and called muonium. This chapter is devoted to the study of the dynamics of muonium, both in vacuum and in a silicon crystal. The positive muon is a spin-1/2 particle which has the same charge as the proton. The muon mass is considerably larger than the electron mass: mµ /me = 206.77. The muon is unstable and decays with a lifetime τ = 2.2 µs. Its use in probing the structure of crystals is based on the rotation of its spin, once a muonium atom is formed: •

It is possible to form muonium atoms in a quantum state such that, at t = 0, the spin state of the µ+ is known. • Using a technique of particle physics, one can measure its spin state at a later time t. • The rotation of the muon spin can be related to the hyperﬁne structure of the 1s level of muonium.

Therefore, the muonium constitutes a local probe, sensitive to electric and magnetic ﬁelds in its vicinity. One can obtain in this way information on the structure of the medium by methods analogous to magnetic resonance experiments. In the ﬁrst part of the chapter, we sketch the principle of the method by studying muonium in vacuum. When the method was ﬁrst applied to a silicon crystal, in 1973, the results seemed anomalous. We shall see in the second section how these results were understood, in 1978, as being due to the anisotropy of crystalline media. Throughout this chapter, the muon will be considered as stable. For simplicity, we set ˆ1 ˆe ≡ µ ˆ2 , ˆ 1 = µ1 σ ˆ 2 = µ2 σ ˆ µ+ ≡ µ µ µ ˆ 1 and σ ˆ 2 are the Pauli matrices. where the (x, y, z) components of σ Numerical values of interest are:

246

24 Probing Matter with Positive Muons

mµ c2 = 105.66 MeV me c2 = 0.511 MeV

µ1 /h = 67.5 MHz T−1 µ2 /h = −1.40 × 104 MHz T−1 .

24.1 Muonium in Vacuum Muonium is formed by slowing down a beam of µ+ , prepared in a given spin state, in a thin metal foil. A suﬃciently slow µ+ can capture an electron and form a hydrogen-like atom in an excited state. This atom falls into its ground state very quickly (in ∼ 10−9 s), the muon’s spin state remaining the same during this process. Once it is formed, the muonium, which is electrically neutral, can diﬀuse outside the metal. We assume that, at t = 0, the state of the muonium atom is the following: • • •

The muon spin is in the eigenstate | + z ≡ |+ of σ ˆ1z . The electron spin is in an arbitrary state α|++β|−, with |α|2 +|β|2 = 1. The wave fuction Ψ (r) of the system is the 1s wave function of the hydrogen-like system, ψ100 (r).

Just as for the hyperﬁne structure of hydrogen, we work in the 4 dimensional Hilbert space corresponding to the spin variables of the electron and the muon. In this Hilbert space, the spin–spin interaction Hamiltonian is A ˆ = E0 − 2 µ0 |ψ100 (0)|2 µ ˆ 2 = E0 + σ ˆ2 , ˆ1 · µ ˆ1 · σ H 3 4π 4 where the indices 1 and 2 refer respectively to the muon and to the electron, and where E0 = −mr c2 α2 /2, with mr being the reduced mass of the (e, µ) system. ˆ in the basis 24.1.1. Write the matrix representation of the Hamiltonian H {|σ1z , σ2z , σiz = ±}. 24.1.2. Knowing the value of A in the hydrogen atom: A/h = 1420 MHz, calculate A in muonium. We recall that µ1 = q¯h/(2mµ ) for the muon, µ2 = h/(2mp ) for the proton, where q is −q¯ h/(2me ) for the electron, µp = 2.79 q¯ the unit charge and mp = 1836.1 me . 24.1.3. Write the general form of an eigenstate of σ ˆ1z with eigenvalue +1: ˆ (i) in the basis {|σ1z , σ2z }; (ii) in the eigenbasis of H. 24.1.4. We assume that, at t = 0, the system is in a state |ψ(0) of the type deﬁned above. Calculate |ψ(t) at a later time. 24.1.5. (a) Show that the operators π ˆ± = (1 ± σ ˆ1z )/2 are the projectors on the eigenstates of σ ˆ1z corresponding to the eigenvalues ±1. (b) Calculate for the state |ψ(t) the probability p(t) that the muon spin is in the state |+ at time t. Write the result in the form

24.2 Muonium in Silicon

247

p(t) = q p+ (t) + (1 − q) p− (t) , where p+ (or p− ) is the probability that one obtains if the electron is initially in the eigenstate of σ2z with eigenvalue +1 (or −1), and where q is a probability, as yet undeﬁned. 24.1.6. In practice, the electronic spins are unpolarized. A rigorous treatment of the problem then requires a statistical description in terms of a density operator. To account for this nonpolarization in a simpler way, we shall set heuristically that the observed probability p¯(t) corresponds to q = 1/2 in the above formula. Using this prescription, give the complete expression for p¯(t).

24.2 Muonium in Silicon We now form muonium in a silicon crystal suﬃciently thick that the muonium does not escape. The muonium stops in an interstitial position inside the crystal lattice, the nearest atoms forming a plane hexagonal mesh around it. The global eﬀect of the interactions between the atoms of the crystal and the muonium atom is to break the spherical symmetry of the spin–spin interaction, but to preserve the rotational symmetry around the z axis perpendicular to the plane of the mesh. We therefore consider the Hamiltonian ˆ = E0 + A σ ˆ2 + D σ ˆ1 · σ ˆ1z σ ˆ2z , H 4

where the constant A may diﬀer from A since the presence of neighboring atoms modiﬁes the Coulomb potential and, therefore, the wave function at the origin. The constants A and D will be determined experimentally; their sign is known: A > 0, D < 0. 24.2.1. Calculate the spin energy levels and the corresponding eigenstates of the muonium trapped in the silicon crystal. 24.2.2. We now reconsider the spin rotation experiment with the following modiﬁcations: • Initially the µ+ spin is now in the eigenstate | + x of σx . • We want to know the probability of ﬁnding the µ+ spin in this same eigenstate | + x at time t. One can proceed as in question 1.5: (a) Calculate in the {|σ1 , σ2 } basis the states |ψ+ (t) and |ψ− (t) which are σ2z is the projection of the electron spin along initially eigenstates of σ ˆ2z (ˆ the z axis). σ1x |ψ (t), where = ±. (b) Evaluate ψ (t)|ˆ

248

24 Probing Matter with Positive Muons

(c) Consider the projector π ˆx = (1 + σ ˆ1x )/2 , and deduce from the previous question the probabilities p± (t). (d) Calculate the measured probability p¯(t) = (p+ (t) + p− (t))/2. 24.2.3. Comparison with experiment: Present day technology in data processing allows one to determine not p(t) itself, but a quantity which is easier to deal with, the characteristic function g(ω) = Re(f (ω)) where ∞ 1 f (ω) = p¯(t) e−t/τ eiωt dt τ 0 is the Fourier transform of p¯(t)e−t/τ /τ . In this expression, the factor e−t/τ /τ is due to the ﬁnite lifetime of the µ+ (τ ∼ 2.2 µs). We recall that ∞ 1 1 . e−t/τ eiωt dt = τ 0 1 − iωτ (a) Figure 24.1a shows the distribution g(ω) as measured in the conditions of question 2.2. Check that this data is compatible with the results found in question 2.2, and deduce from the data the values of A /h and D/h (we recall that D < 0). g(ω)

37,25

54,85

(arbitrary units) (a)

0

10

100

50

(

)

(

)

ω/2π MHz

g(ω) (b) 92,1

0

10

50

100

ω/2π MHz

Fig. 24.1. Experimental variations of the quantity g(ω), deﬁned in the text, with the frequency ν = ω/(2π). (a) In the conditions described in question 2.2, and (b) in another experimental conﬁguration

24.3 Solutions

249

(b) Figure 24.1b is obtained by a slight modiﬁcation of the previous experiment. Can you tell what modiﬁcation has been made? How can one evaluate the position of the third peak, in terms of the constants of the problem?

24.3 Solutions Section 24.1: Muonium in Vacuum 24.1.1. The Hamiltonian is ˆ = E0 + A (ˆ H σ1x σ ˆ2x + σ ˆ1y σ ˆ2y + σ ˆ1z σ ˆ2z ) . 4 The matrix representation is therefore ⎛ ⎞ E0 + A/4 0 0 0 ⎜ ⎟ A/2 0 0 E0 − A/4 ˆ =⎜ ⎟ , H ⎝ ⎠ 0 0 A/2 E0 − A/4 0 0 0 E0 + A/4 where the elements are ordered as: | + +, | + −, | − +, | − −. 24.1.2. The constant A is related to its value in the hydrogen atom by A |ψ(0)|2 µ1 |ψ(0)|2 mp 1 . = = 2 AH |ψ(0)|H µp |ψ(0)|2H mµ 2.79 In ﬁrst approximation, muonium and hydrogen have similar sizes and wave functions, since the muon is much heavier than the electron. Therefore we obtain A AH (mp /2.79 mµ ) and A/h 4519 MHz. The reduced mass correction to the value of the wave function at the origin is straightforward to calculate. It is of the order of 1% and it leads to A = 4519 (1 − 0.0126) = 4462 MHz . h This value is very close to the observed 4463 MHz, the diﬀerence being due to relativistic eﬀects. 24.1.3. The state under consideration can be written as |ψ = |+ ⊗ (α|+ + β|−)

with |α|2 + |β|2 = 1 .

Equivalently, one can write it as |ψ = α| + + + β| + −. ˆ consists in the common eigenstates of the total spin The eigenbasis of H 2 ˆ ˆ operators S and Sz : ⎧ ⎨ | + + √ triplet states (| + − + | − +)/ 2 ⎩ | − − √ singlet state (| + − − | − +)/ 2 .

250

24 Probing Matter with Positive Muons

Therefore, one also has the representation β |ψ = α|1, 1 + √ (|1, 0 + |0, 0) , 2 where the only constraint on α and β is |α|2 + |β|2 = 1. 24.1.4. We start from |ψ(0) = |ψ as deﬁned above. The energy levels and the corresponding eigenstates are known: triplet states

ET = E0 + A/4

singlet state

ES = E0 − 3A/4 .

At time t the state is:

β i3At/4¯h β −iE0 t/¯ h −iAt/4¯ h |0, 0 . |ψ(t) = e e α |1, 1 + √ |1, 0 + √ e 2 2 24.1.5. (a) It is straightforward to check that π ˆ± are projectors: ˆ+ |− = 0 π ˆ+ |+ = |+ π ˆ− |+ = 0 . π ˆ− |− = |− π (b) The probability of ﬁnding the muon spin in the state |+ at time t is by deﬁnition p(t) = ˆ π+ |ψ(t)2 = ψ(t)|ˆ π+ |ψ(t) . Using π ˆ+ |1, 1 = |1, 1 1 ˆ+ |0, 0 = √ | + − π ˆ+ |1, 0 = π 2 π ˆ+ |1, −1 = 0 , we obtain β π ˆ+ |ψ(t) = e−i(E0 +A/4)t/¯h α| + + + 1 + eiAt/¯h | + − . 2 Squaring the norm of π ˆ+ |ψ(t), we get h)) . p(t) = |α|2 + |β|2 cos2 (At/(2¯ There is a periodic modulation of the probability of observing the muon spin aligned with the positive z axis, which can be interpreted as a rotation of the muon spin with frequency ν = A/h.

24.3 Solutions

251

• The probability p+ (t) corresponds to the initial state |ψ(0) = |++. This is a stationary state so that p(t) ≡ p+ (t) = 1 in this case. to the initial state |ψ(0) = • The probability p− (t) corresponds √ | + − = (|1, 0 + |0, 0)/ 2. There is in this case an oscillation with a 100 % modulation between | + − and | − +, so that p(t) ≡ p− (t) = h)). cos2 (At/2(¯ Therefore the result can be cast in the form suggested in the text: p(t) = q p+ (t) + (1 − q) p− (t) , with q = |α|2 . 24.1.6. When the electronic spins are unpolarized, we obtain following the assumption of the text: p¯(t) =

3 1 + cos(At/¯h) . 4 4

Note: The rigorous way to treat partially polarized systems is based on the density operator formalism. In the present case the density operator for the unpolarized electron is: ρ2 =

1 (|++| + |−−|) , 2

so that the initial density operator for the muonium is: 1 1 | + ++ + | + | + −+ − | 2 2 1 = |1, 11, 1| 2 1 + (|1, 01, 0| + |1, 00, 0| + |0, 01, 0| + |0, 00, 0|) . 4

ρ(0) =

The density operator at time t is then given by: ρ(t) =

1 |1, 11, 1| 2 1 + |1, 01, 0| + e−iAt/¯h |1, 00, 0| 4 +eiAt/¯h |0, 01, 0| + |0, 00, 0|

hence the probability: p¯(t) = +, +|ρ(t)|+, + + + − |ρ(t)| + −

1 1 1 1 1 1 = + + e−iAt/¯h + eiAt/¯h + 2 4 2 2 2 2 3 1 h) . = + cos(At/¯ 4 4

252

24 Probing Matter with Positive Muons

Section 24.2: Muonium in Silicon 24.2.1. In the factorized basis {|σ1 , σ2 }, the Hamiltonian is written as ⎞ ⎛ 0 0 0 A /4 + D ⎟ ⎜ A /2 0 0 −A /4 − D ˆ = E0 + ⎜ ⎟ . H ⎠ ⎝ −A /4 − D 0 0 A /2 0 0 0 A /4 + D This Hamiltonian is diagonal in the eigenbasis {|S, m} of the total spin. A simple calculation shows that the eigenvalues and eigenvectors are E1 = E4 = E0 + A /4 + D E2 = E0 + A /4 − D E3 = E0 − 3A /4 − D

|1, 1 and |1, −1 |1, 0 |0, 0 .

24.2.2. (a) The initial states |ψ+ (0) and |ψ− (0) are easily obtained in the factorized basis as √ |ψ+ (0) = | + x ⊗ |+ = (| + + + | − +)/√2 |ψ− (0) = | + x ⊗ |− = (| + − + | − −)/ 2 . They can be written in the total spin basis {|S, m} as 1 1 |ψ+ (0) = √ |1, 1 + (|1, 0 − |0, 0) 2 2 1 1 |ψ− (0) = √ |1, −1 + (|1, 0 + |0, 0) . 2 2 h, we ﬁnd at time t: Writing ωi = −Ei /¯ eiω3 t eiω2 t eiω1 t |ψ+ (0) = √ |1, 1 + |1, 0 − |0, 0 2 2 2 eiω3 t eiω2 t eiω4 t |ψ− (0) = √ |1, −1 + |1, 0 + |0, 0 , 2 2 2 which can now be converted in the factorized basis: eiω2 t − eiω3 t eiω2 t + eiω3 t eiω1 t √ √ |ψ+ (t) = √ | + + + | + − + | − + 2 2 2 2 2 eiω2 t + eiω3 t eiω2 t − eiω3 t eiω4 t √ √ |ψ− (t) = √ | − − + | + − + | − + . 2 2 2 2 2

24.3 Solutions

253

(b) Since σ ˆ1x |σ1 , σ2 = | − σ1 , σ2 , the matrix elements ψ± (t)|ˆ σ1x |ψ± (t) are equal to:

1 Re e−iω1 t eiω2 t + eiω3 t 2

(A + 2D)t 1 2Dt + cos = cos 2 h ¯ ¯h " −iω4 t iω2 t

# 1 e σ1x |ψ− (t) = Re e + eiω3 t . ψ− (t)|ˆ 2 ψ+ (t)|ˆ σ1x |ψ+ (t) =

Since ω1 = ω4 , the two quantities are equal. (c) The desired probabilities are p± (t) = ˆ π+x |ψ± (t)2 = ψ± (t)|ˆ π+x |ψ± (t) or, equivalently, 1 1 1 ˆ1x ) |ψ± (t) = + ψ± (t)|ˆ p± (t) = ψ± (t)| (1 + σ σ1x |ψ± (t) . 2 2 2 Using the result obtained above, we get:

(A + 2D)t 1 1 2Dt + cos p± (t) = + cos . 2 4 h ¯ ¯h (d) Since p+ (t) = p− (t), the result for p¯(t) is simply:

(A + 2D)t 1 1 2Dt + cos p¯(t) = + cos . 2 4 h ¯ ¯h 24.2.3. Comparison with Experiment: In practice, the time t corresponds to the decay of the µ+ , with the emission of a positron e+ and two neutrinos. The positron is suﬃciently energetic and leaves the crystal. It is emitted preferentially in the muon spin direction. One therefore measures the direction where the positron is emitted as a function of time. For N0 incoming muons, the number of positrons emitted in the x direction is dN (t) = N0 p¯(t)e−t/τ dt/τ , where τ is the muon lifetime. A convenient way to analyse the signal, and to extract the desired frequencies, consists in taking the Fourier transform of the above signal. Deﬁning ∞ 1 1 , e(iω−1/τ )t dt = f0 (ω) = τ 0 1 − iωτ one obtains

1 1 2D 2D f (ω) = f0 (ω) + f0 ω − + f0 ω + 2 8 h ¯ ¯h

1 A + 2D A + 2D + f0 ω − + f0 ω + . 8 h ¯ ¯h

254

24 Probing Matter with Positive Muons

The function Re(f0 (ω)) has a peak at ω = 0 whose half width is 1/τ , which corresponds to 100 kHz. (a) The curve of Fig. 24.1 is consistent with this observation. Besides the h. peak at ω = 0, we ﬁnd two peaks at ω1 = −2D/¯h and ω2 = (A + 2D)/¯ Assuming that D is negative, which can be conﬁrmed by a more thorough analysis, one obtains 2D/h = −37.25 MHz

and

A /h = 92.1 MHz .

(b) In general, one expects to see peaks at all frequencies ωi − ωj , and in h. In order to observe the corresponding peak, one particular at ω2 −ω3 = −A /¯ must measure the µ+ spin projection along a direction which is not orthogonal to the z axis. This leads to a term in cos (ω2 − ω3 )t in p¯(t), which appears in Fig. 24.1.

25 Quantum Reﬂection of Atoms from a Surface

This chapter deals with the reﬂection of very slow hydrogen atoms from a surface of liquid helium. In particular, we estimate the sticking probability of the atoms onto the surface. This sticking proceeds via the excitation of a surface wave, called a ripplon. We show that this probability must vanish at low temperatures, and that, in this limit, the reﬂection of the atoms on the surface is specular. In all the chapter, the position of a particle is deﬁned by its coordinates r = (x, y) in a horizontal plane, and its altitude z. The altitude z = 0 represents the position of the surface of the liquid He bath at rest. The wave functions ψ(r, z) of the H atoms are normalized in a rectangular box of volume Lx Ly Lz . We write m for the mass of a H atom (m = 1.67 10−27 kg).

25.1 The Hydrogen Atom–Liquid Helium Interaction Consider a H atom above a liquid He bath at rest (cf Fig. 25.1). We model the H-liquid He interaction as the sum of pairwise interactions between the H atom at point (R, Z) (Z > 0), and the He atoms at (r, z), with z < 0: +∞ dz U ( (R − r)2 + (Z − z)2 ) Θ(−z) V0 (Z) = n d2 r −∞

where n is the number of He atoms per unit volume, and Θ is the Heaviside function. 25.1.1. We recall the form of the Van der Waals potential: U (d) = −

C6 , d6

which describes the long distance interaction between a H atom and a He atom separated by a distance d. Show that the long distance potential between the H atom and the liquid He bath is of the form:

256

25 Quantum Reﬂection of Atoms from a Surface

Fig. 25.1. A hydrogen atom incident on a liquid helium bath. The oscillations of the surface of the liquid are studied in Sect. 25.2

V0 (Z) = −

α . Z3

Express α in terms of C6 and n. 25.1.2. Experimentally, one ﬁnds α = 1.9 × 10−2 eV ˚ A3 . At what distance from the surface does gravity become larger than the Van der Waals force? In what follows, we shall neglect the gravitational force. 25.1.3. Show that the eigenstates of the Hamiltonian which describes the motion of the H atom are of the form |k⊥ , φσ , where k⊥ represents a plane running wave propagating in the plane Oxy, i.e. parallel to the surface of the liquid He, and where φσ is an eigenstate of the Hamiltonian which describes the motion along the z axis: R, Z|k⊥ , φσ =

1 ei(kx X+ky Y ) φσ (Z) . Lx Ly

25.1.4. We want to evaluate the number of bound states of the motion along the z axis in the potential: α Z3 V0 (Z) = +∞ V0 (Z) = −

if Z > zmin if Z ≤ zmin .

We shall use the WKB approximation. (a) Justify the shape of this potential. (b) What is the continuity condition for the wave function at Z = zmin ? (c) Show that the quantization condition for a motion with turning points zmin and b is

b 3 k(Z) dZ = n + π 4 zmin with n integer ≥ 0.

25.2 Excitations on the Surface of Liquid Helium

257

(d) Infer the order of magnitude of the number of bound states as a function of zmin and α. What is the domain of validity of the result? A. (e) The parameter zmin for the surface of liquid He is of the order of 2 ˚ How many bound states does one expect for the motion along the z axis? (f) Experimentally, one ﬁnds that there is a single bound state H–liquid He, whose energy is E0 = −8.6 × 10−5 eV. Compare this result with the WKB prediction. This unique bound state in the z-axis motion will be denoted φ0 in the rest of the chapter.

25.2 Excitations on the Surface of Liquid Helium The general dispersion relation for waves propagating on the surface of a liquid is A ωq2 = gq + q 3 with q = |q| , ρ0 where ωq and q = (qx , qy ) are, respectively, the frequency and the wave vector of the surface wave, g is the acceleration of gravity and A and ρ0 represent the surface tension and the mass density of the liquid. 25.2.1. Discuss the nature of the surface waves (capillary waves or gravity waves) according to the value of the wavelength λ = 2π/q. Perform the numerical application in the case of liquid He: ρ0 = 145 kg m−3 , A = 3.5 × 10−4 Jm−2 . 25.2.2. Hereafter, we are only interested in waves for which h ¯ ωq |E0 |. Show that these are always capillary waves and give their wavelengths. In what follows we shall use the simpler dispersion relation ωq2 = (A/ρ0 )q 3 . 25.2.3. In order to quantize these surface waves, we introduce the bosonic operators rq and rq† corresponding to the annihilation and the creation of an excitation quantum. These elementary excitations are called ripplons. The Hamiltonian which describes these excitations is:1 HS =

q max q

hωq rˆq† rˆq . ¯

The altitude h(r) of the liquid surface at point r = (x, y) becomes a twodimensional scalar ﬁeld operator: ˆ h(r) =

q max q

1

hq (rq† e−iq.r + rq eiq.r )

with

hq =

¯q h . 2ρ0 ωq Lx Ly

The summation over q is limited to q < qmax where qmax is of the order of a fraction of an inverse ˚ Angstrom. For larger values of q, hence smaller wavelengths, the description of the vicinity of the surface in terms of a ﬂuid does not hold any longer.

258

25 Quantum Reﬂection of Atoms from a Surface

Evaluate, at zero temperature, the r.m.s. altitude ∆h of the position of the surface. We recall that, in two dimensions, the conversion of a discrete summation into an integral proceeds via: Lx Ly −→ d2 q . 2 4π q A−1 . Numerical application: qmax = 0.5 ˚

25.3 Quantum Interaction Between H and Liquid He We now investigate the modiﬁcations to the H–liquid He potential arising from the possible motion of the surface of the liquid helium bath. In order to do so, we replace the coupling considered above by +∞ ˆ dz U ( (R − r)2 + (Z − z)2 ) Θ(h(r) − z) . V (R, Z) = n d2 r −∞

ˆ and interpret the result. 25.3.1. Expand V (R, Z) to ﬁrst order in h ˆ 25.3.2. Replacing h(r) by its expansion in terms of operators rˆq , rˆq† , cast V (R, Z) in the form:

hq e−iq.R Vq (Z)rq† + h.c. V (R, Z) = V0 (Z) + q

with Vq (Z) = n

d2 r e−iq.r U ( r 2 + Z 2 ).

25.3.3. Introducing the creation operators a ˆ†k,σ and the annihilation operators a ˆk,σ of a hydrogen atom in an eigenstate of the motion in the potential V0 (Z), write in second quantization the total hydrogen–ripplon Hamiltonian ˆ to ﬁrst order in h.

25.4 The Sticking Probability We consider a H atom in an asymptotically free state in the z direction (i.e. behaving as e±ikσ z as z → +∞). This state denoted |k⊥ , φσ has an energy Ei =

¯2 2 h (k + kσ2 ) . 2m ⊥

We now calculate the probability that this atom sticks on the surface, which is assumed here at zero temperature.

25.5 Solutions

259

25.4.1. How does the matrix element φ0 |Vq |φσ vary with the size of the normalization box? We assume in the following that this matrix element is proportional to kσ if kσ is suﬃciently small, and we introduce M (q) such that φ0 |Vq |φσ = √

hkσ ¯ M (q) . 2mLz

All following results will be expressed in terms of M (q). 25.4.2. Using Fermi’s Golden Rule, deﬁne a probability per unit time for an atom to stick on the surface. In order to do so, one will deﬁne properly: (a) the continuum of ﬁnal states; (b) the conditions imposed by energy conservation. For simplicity, we shall assume that the incident energy Ei is negligible compared to the bound state energy E0 . Show that the emitted ripplon has a wave vector q such that |q| = q0 with:

h ¯

A 3/2 h ¯ 2 q02 = |E0 | ; q0 + ρ0 r2m

(c) the density of ﬁnal states. 25.4.3. Express the ﬂux of incident atoms in terms of ¯h, kσ , m and Lz . 25.4.4. Write the expression for the probability that the hydrogen atom sticks on the surface of the liquid helium bath in terms of ¯h, q0 , A, ρ0 , kσ and M (q). Check that this probability is independent of the normalization volume Lx Ly Lz . 25.4.5. How does this probability vary with the energy of the incident hydrogen atoms? 25.4.6. Describe qualitatively how one should modify the above treatment if the liquid helium bath is not at zero temperature.

25.5 Solutions Section 25.1: The Hydrogen Atom–Liquid Helium Interaction 25.1.1. We use cylindrical coordinates, assuming that the H atom is at R = 0. The potential V0 (Z) takes the form

260

25 Quantum Reﬂection of Atoms from a Surface

V0 (Z) = n

0

d2 r −∞

= −nC6

dz U ( r2 + (Z − z)2 )

0

∞

dz −∞ 0

π = − nC6 2

dr 0

−∞

2πr (r2 + (Z − z)2 )3

dz πnC6 =− . 4 (Z − z) 6Z 3

Therefore

α πnC6 . with α= Z3 6 25.1.2. The force which derives from V0 (Z) has modulus V0 (Z) = −

F (Z) =

3α . Z4

We have 3α/Zg4 = M g for Zg = (3α/(M g))1/4 . The numerical application yields Zg = 0.86 µm which is very large on the atomic scale. For all the relevant H–liquid He distances, which are between 0.1 nm and 1 nm, gravity can be neglected. ˆ Z , where ˆ =H ˆ⊥ + H 25.1.3. The Hamiltonian can be split as H 2 pˆ2 ˆ ⊥ = pˆx + y H 2m 2m

2 ˆ Z = pˆz + V0 (Z) ˆ . H 2m

and

These two Hamiltonians commute and the eigenbasis of the total Hamiltonian ˆ ⊥ , where k⊥ ˆ is factorized as a product |k⊥ , φσ of (i) the eigenstates of H H represents the wave vector of a plane running wave propagating in the (x, y) ˆ Z which describes the motion along the plane, and (ii) the eigenstates φσ of H z axis. 25.1.4. (a) For Z ≤ zmin , the overlap of the electron wave functions of the H and He atoms causes a repulsion between these atoms, which is modeled here by a hard core potential. For Z zmin , the Van der Waals forces are dominant. (b) For Z ≤ zmin , the wave function φ(Z) is such that φ(Z) = 0. Since φ(Z) is continuous, we have φ(zmin ) = 0. (c) For a turning point b, the WKB eigenfunction of energy E has the following form in the allowed region (E > V0 (Z)): φ(Z) =

C k(Z)

b

cos Z

π k(Z ) dZ − 4

where C is a normalization constant and where

,

25.5 Solutions

hk(Z) = ¯

261

2m(E − V0 (Z)) .

Imposing the condition φ(zmin ) = 0 yields

b

π k(Z ) dZ − = 4 zmin

1 n+ 2

3 k(Z ) dZ = n + π 4 zmin

π , i.e.

b

with n a positive integer. (d) If the WKB method were exact, the number of bound states would be

∞ k(Z ) 3 dZ − n = 1 + Int , π 4 zmin where Int denotes the integer part and where k(Z) is calculated for a zero energy E. As usual for the WKB method, the accuracy of this expression is good if is large. We can take in this case: ∞the number of bound states ¯ k(Z) = 2mα/Z 3 , which yields n π −1 zmin k(Z ) dZ with h 2mα 2 . n π¯ h zmin (e) The above formula yields n 1.36. We therefore expect a number of bound states close to 1, say between 0 and 2. (f ) The experimental result compares favorably with the WKB. prediction, but it is beyond the validity of the WKB approximation to give a correct expression for φ0 (Z). Section 25.2: Excitations on the Surface of Liquid Helium 25.2.1. The two terms of the dispersion relation are equal if q = gρ0 /A or, equivalently, for a wavelength A . gρ0

λ = 2π

Numerically, one obtains λ = 3 mm. Therefore, we observe capillary waves (ωq2 Aq 3 /ρ0 ) for λ 3 mm, and gravity waves (ωq2 gq) for λ 3 mm. For λ = 3 mm the corresponding energy is ¯hωq = 1.3 × 10−13 eV. 25.2.2. For an energy such that |E0 | 10−13 eV, we are therefore in the regime of capillary waves, with the wavelength: 2π = 2π λ= q The numerical value is λ = 33 ˚ A.

Ah2 ρ0 E02

1/3 .

262

25 Quantum Reﬂection of Atoms from a Surface

25.2.3. We have ˆ 2 − h ˆ 2 = h ˆ 2 = ∆h2 = h

h2q rq rq† =

q

h2q .

q

Converting this into an integral, we obtain qmax Lx Ly ¯h hq ¯ √ 2 √ ∆h2 = d q = q dq 2 4π 2ρ0 ωq Lx Ly 4π Aρ0 0 =

¯ h 6π

3 qmax hωmax ¯ , = Aρ0 6πA

which yields ∆h = 0.94 ˚ A. Section 25.3: Quantum Interaction Between H and Liquid He ˆ 25.3.1. Using the fact that Θ (z) = δ(z), we can write Θ(−z + h(r)) ˆ Θ(−z) + h(r)δ(z) since the δ function is even. Therefore, we obtain ˆ . V (R, Z) V0 (Z) + n d2 r U ( (R − r)2 + Z 2 ) h(r) In this expression, the second term describes the interaction with the “additional” or “missing” atoms on the surface as compared to the equilibrium position z = 0. ˆ 25.3.2. Replacing h(r) by its expansion we obtain V (R, Z) V0 (Z) + n d2 r U ( (R − r)2 + Z 2 ) hq (ˆ rq† e−iq·r + rˆq eiq·r ) . q

Considering the term rq† and setting r = r−R, we obtain in a straightforward manner

hq e−iq·R Vq (Z)ˆ rq† + h.c. , V (R, Z) = V0 (Z) + q

with Vq (Z) = n

d2 r e−iq·r U ( r 2 + Z 2 ) .

ˆ at = 25.3.3. The Hamiltonian is the sum of the “free” Hamiltonians H ˆ S , and the coupling term found above. One has V0 (Z) and H ˆS = ˆ at = H H Ek,σ a ˆ†k,σ a ˆk,σ ¯hωq rˆq† rˆq . k,σ

The coupling term becomes

q

P2 2M

+

25.5 Solutions

k,σ

k ,σ

q

263

hq a ˆ†k,σ a ˆk ,σ rˆq† k, φσ |e−iq.R Vq (Z)|k , φσ + h.c.

The matrix element is k, φσ |e−iq·R Vq (Z)|k , φσ = k|e−iq·R |k φσ |Vq (Z)|φσ = δk ,k+q φσ |Vq (Z)|φσ . ˆ We end up with the total hydrogen–ripplon Hamiltonian to ﬁrst order in h: ˆ = H Ek,σ a ˆ†k,σ a ˆk,σ + hωq rˆq† rq ¯ q

k,σ

+

q,k,σ,σ

hq a ˆ†k,σ a ˆk+q,σ rˆq†

φσ |Vq (Z)|φσ + h.c. .

In the (x, y) plane, the momentum is conserved owing to the translation invariance of the problem. This can be seen directly on the form of the coupling a ˆ†k,σ a ˆk+q,σ rˆq† , which annihilates a H atom with momentum h ¯ (k + q) term, and creates a H atom with momentum h ¯ k and a ripplon with momentum h ¯ q. Section 25.4: The Sticking Probability 25.4.1. We have by deﬁnition φ0 |Vq |φσ =

φ∗0 (Z) Vq (Z) φσ (Z) dZ .

Since |φσ is an asymptotically free state, it is normalized in a segment of −1/2 length Lz . Therefore its amplitude varies as Lz . Since |φ0 is a localized state which does not depend on Lz , we ﬁnd 1 φ0 |Vq |φσ ∝ √ . Lz The fact that this matrix element is proportional to kσ in the limit of small incident momenta is more subtle. The positions Z contributing to the matrix element are close to zero, since the bound state φ0 (Z) is localized in the vicinity of the He surface. Therefore only the values of φσ (Z) around Z = 0 are relevant for the calculation of the integral. For the Z −3 potential between the H atom and the He surface, one ﬁnds that the amplitude of φσ in this region is proportional to kσ , hence the result. Such a linear dependance can be recovered analytically by replacing the Z −3 potential by a square well, but is out √ of reach of the WKB approximation, which would predict a dependance in kσ for the amplitude around Z = 0 of φσ . The reason for this discrepancy is that the potential in −αZ −3 is too stiﬀ for the WKB to be valid for the calculation of φσ at distances larger than mα/¯h2 .

264

25 Quantum Reﬂection of Atoms from a Surface

25.4.2. We start with the initial state k⊥ , φσ . If the atom sticks to the surface, the ﬁnal state along the z axis is |φ0 . The sticking proceeds via the emission of a ripplon of momentum h ¯ q and a change of the transverse mohq. mentum h ¯ k⊥ → ¯hk⊥ − ¯ (a) The continuum of ﬁnal states is characterized by the vector q: |k⊥ , φσ → |k⊥ − q, φ0 ⊗ |q .

(b) Energy conservation gives Ei = Ef with:

Ei =

2 ) ¯ 2 (kσ2 + k⊥ h 2m

Ef = E0 +

¯ 2 (k⊥ − q)2 h +h ¯ ωq . 2m

We suppose that Ei is negligible compared to the bound state energy E0 . 2 /(2m) is also very small compared to Therefore ¯h2 (k⊥ · q)/m ∼ |E0 |¯ h2 k⊥ |E0 |, and we obtain: ¯ 2 q2 h +h ¯ ωq |E0 | . 2m This equation, in addition to the dispersion relation for ripplons, determines the modulus q0 of q: h ¯

A 3/2 h ¯ 2 q02 = |E0 | . q0 + ρ0 2m

(c) A variation δE of the ﬁnal state energy corresponds to a variation δq such that 3¯ h Aq0 h 2 q0 ¯ + δq = δE . m 2 ρ0 The number of states δ 2 n in a domain δ 2 q is:

δ2 n =

Lx Ly 2 Lx Ly δ q= q0 δq δθ . 2 4π 4π 2

After integrating over δθ, we obtain:

ρ(Ef ) =

mq0 Lx Ly . 2 π 2¯ h q0 + 3m¯h Aq0 /ρ0

25.6 Comments

265

25.4.3. The number of atoms which cross a plane of altitude Z in the direc¯ kσ dt/(2mLz ). The ﬂux tion Z < 0 during a time interval dt is vz dt/(2Lz ) = h is therefore: hkσ ¯ . Φσ = 2mLz 25.4.4. The sticking probability is the ratio of the probability per unit time, given by Fermi’s Golden Rule, and the incident ﬂux: P =

2π 2mLz |k⊥ , φσ |V |k⊥ − q, φ0 , q|2 ρ(Ef ) . h ¯ ¯hkσ

This reduces to P =

mkσ |M (q)|2 √ . 3Am + 2¯ h Aρ0 q0

√ 25.4.5. P varies as kσ ∝ E. At very small energies, the sticking probability goes to zero and the H atoms bounce elastically on the liquid He surface. 25.4.6. If the liquid helium bath is not at zero temperature, other processes can occur, in particular a sticking process accompanied by the stimulated emission of a ripplon. One must therefore take into account the number nq0 of thermal ripplons.

25.6 Comments The theory of the sticking of H atoms onto a surface of liquid He can be found in the [1] below. Thorough experimental studies of this process are presented in [2] and [3]. 1. D.S. Zimmerman and A.J. Berlinsky, Can. J. Phys. 61, 508 (1983). 2. J.J. Berkhout, O.J. Luiten, J.D. Setija, T.W. Hijmans, T. Mizusaki, and J.T.M. Walraven, Phys. Rev. Lett. 63, 1689 (1989). 3. J.M. Doyle, J.C. Sandberg, I.A. Yu, C.L. Cesar, D. Kleppner, and T.J. Greytak, Phys. Rev. Lett. 67, 603 (1991).

26 Laser Cooling and Trapping

By shining laser light onto an assembly of neutral atoms or ions, it is possible to cool and trap these particles. In this chapter we study a simple cooling mechanism, Doppler cooling, and we derive the corresponding equilibrium temperature. We then show that the cooled atoms can be conﬁned in the potential well created by a focused laser beam. We consider a “two state” atom, whose levels are denoted |g (ground state) and |e (excited state), with respective energies 0 and ¯hω0 . This atom interacts with a classical electromagnetic wave of frequency ωL /2π. For an atom located at r, the Hamiltonian is ˆ =h H ¯ ω0 |ee| − d · (E(r, t)|eg| + E ∗ (r, t)|ge|) ,

(26.1)

where d, which is assumed to be real, represents the matrix element of the atomic electric dipole operator between the states |g and |e (i.e. d = ˆ ˆ ∗ ). The quantity E + E ∗ represents the electric ﬁeld. We e|D|g = g|D|e set E(r, t) = E 0 (r) exp(−iωL t) . In all the chapter we assume that the detuning ∆ = ωL −ω0 is small compared with ωL and ω0 . We treat classically the motion r(t) of the atomic center of mass.

26.1 Optical Bloch Equations for an Atom at Rest 26.1.1. Write the evolution equations for the four components of the density operator of the atom ρgg , ρeg , ρge and ρee under the eﬀect of the Hamiltonian ˆ H. 26.1.2. We take into account the coupling of the atom with the empty modes of the radiation ﬁeld, which are in particular responsible for the spontaneous emission of the atom when it is in the excited state |e. We shall assume that this boils down to adding to the above evolution equations “relaxation” terms:

268

26 Laser Cooling and Trapping

d d ρee ρgg =− = −Γ ρee dt dt relax relax

d d Γ Γ ρeg ρge = − ρeg = − ρge , dt 2 dt 2 relax relax

where Γ −1 is the radiative lifetime of the excited state. Justify qualitatively these terms. 26.1.3. Check that for times much larger than Γ −1 , these equations have the following stationary solutions: s 2(s + 1) 2+s = 2(s + 1)

d · E(r, t)/¯ h 1 ∆ + iΓ/2 1 + s h 1 d · E ∗ (r, t)/¯ =− ∆ − iΓ/2 1 + s

ρee =

ρeg = −

ρgg

ρge

where we have set s=

h2 2 |d · E 0 (r)|2 /¯ . 2 2 ∆ + Γ /4

26.1.4. Interpret physically the steady state value of the quantity Γ ρee in terms of spontaneous emission rate.

26.2 The Radiation Pressure Force In this section, we limit ourselves to the case where the electromagnetic ﬁeld is a progressive plane wave: E 0 (r) = E 0 exp(ik · r) . By analogy with the classical situation, we can deﬁne the radiative force operator at point r as: ˆ (r) = −∇r H ˆ . F ˆ (r) assuming that the atom is at 26.2.1. Evaluate the expectation value of F rest in r and that its internal dynamics is in steady state. 26.2.2. Interpret the result physically in terms of momentum exchanges between the atom and the radiation ﬁeld. One can introduce the recoil velocity vrec = h ¯ k/m. 26.2.3. How does this force behave at high intensities? Give an order of magnitude of the possible acceleration for a sodium atom 23 Na, with a resonance wavelength λ = 0.589 × 10−6 m and a lifetime of the excited state of Γ −1 = 16 × 10−9 s (d = 2.1 × 10−29 C m). 26.2.4. We now consider an atom in uniform motion: r = r 0 + v 0 t (v0 c). Give the expression for the force acting on this atom.

26.3 Doppler Cooling

269

26.2.5. The action of the force on the atom will modify its velocity. Under what condition is it legitimate to treat this velocity as a constant quantity for the calculation of the radiation pressure force, as done above? Is this condition valid for sodium atoms?

26.3 Doppler Cooling The atom now moves in the ﬁeld of two progressive plane waves of opposite directions (+z and −z) and of same intensity (Fig. 26.1). We restrict ourselves to the motion along the direction of propagation of the two waves and we assume that for weak intensities (s 1) one can add independently the forces exerted by the two waves.

Fig. 26.1. Doppler cooling in one dimension

26.3.1. Show that for suﬃciently small velocities, the total force is linear in the velocity and can be cast in the form: f =−

mv . τ

26.3.2. What is the minimal (positive) value of τmin for a ﬁxed saturation parameter per wave s0 for an atom at rest? Calculate τmin for sodium atoms, assuming one ﬁxes s0 = 0.1. 26.3.3. This cooling mechanism is limited by the heating due to the random nature of spontaneous emission. To evaluate the evolution of the velocity distribution P (v, t) and ﬁnd its steady state value, we shall proceed in the following way: (a) Express P (v, t + dt) in terms of P (v, t). One will split the atoms into three classes: • the atoms having undergone no photon scattering event between t and t + dt, • the atoms having scattered a photon from the +z wave, • the atoms having scattered a photon from the −z wave. We choose dt short enough that the probability of the ﬁrst option is dominant, and such that multiple scattering events are negligible. We also assume that the velocities contributing signiﬁcantly to P (v, t) are

270

26 Laser Cooling and Trapping

small enough for the linearization of the force performed above to be valid. For simplicity we will assume that spontaneously emitted photons propagate only along the z axis, a spontaneous emission occurring with equal probabilities in the directions +z and −z. (b) Show that P (v, t) obeys the Fokker–Planck equation ∂ ∂2P ∂P = α (vP ) + β 2 ∂t ∂v ∂v and express of α and β in terms of the physical parameters of the problem. (c) Determine the steady state velocity distribution. Show that it corresponds to a Maxwell distribution and give the eﬀective temperature. (d) For which detuning is the eﬀective temperature minimal? What is this minimal temperature for sodium atoms?

26.4 The Dipole Force We now consider a stationary light wave (with a constant phase) E 0 (r) = E ∗0 (r) . ˆ (r) = 26.4.1. Evaluate the expectation value of the radiative force operator F ˆ −∇r H assuming that the atom is at rest in r and that its internal dynamics has reached its steady state. 26.4.2. Show that this force derives from a potential and evaluate the potential well depth that can be attained for sodium atoms with a laser beam of intensity P = 1 W, focused on a circular spot of radius 10 µm, and a wavelength λL = 0.650 µm.

26.5 Solutions Section 26.1: Optical Bloch Equations for an Atom at Rest 26.1.1. The evolution of the density operator ρˆ is given by: i¯ h

dˆ ρ ˆ ρˆ] = [H, dt

so that: dρee d · E(r) e−iωL t d · E ∗ (r) eiωL t =i ρge − i ρeg dt h ¯ ¯h dρeg d · E(r) e−iωL t = −iω0 ρeg + i (ρgg − ρee ) dt h ¯

26.5 Solutions

and dρee dρgg =− dt dt

dρge = dt

dρeg dt

271

∗ .

26.1.2. Assume that the atom-ﬁeld system is placed at time t = 0 in the state |ψ(0) = (α|g + β|e) ⊗ |0 , where |0 denotes the vacuum state of the electromagnetic ﬁeld and neglect in a ﬁrst step the action of the laser. At time t, the state of the system is derived from the Wigner–Weisskopf treatment of spontaneous emission: |ψ(t) = (α|g + βe−(iω0 +Γ/2)t) |e) ⊗ |0 + |g ⊗ |φ , where the state of the ﬁeld |φ is a superposition of one-photon states for the various modes of the electromagnetic ﬁeld. Consequently the evolution of the density matrix elements is ρee (t) = |β|2 e−Γ t , ρeg (t) = α∗ βe−(iω0 +Γ )t , or, in other words,

dρeg dρee Γ = −Γ ρee = − ρeg . dt relax dt relax 2 The two other relations originate from the conservation of the trace of the density operator (ρee + ρgg = 1) and from its hermitian character ρeg = ρ∗ge . We assume in the following that the evolution of the atomic density operator is obtained by adding the action of the laser ﬁeld and the spontaneous emission contribution. Since Γ varies like ω03 , this is valid as long as the shift of the atomic transition due to the laser irradiation remains small compared hω0 , which is satisﬁed for usual continuous laser with ω0 . This requires dE ¯ sources. 26.1.3. The evolution of the density operator components is given by d · E(r) e−iωL t d · E ∗ (r) eiωL t dρee = −Γ ρee + i ρge − i ρeg dt h ¯ ¯h

dρeg Γ d · E(r) e−iωL t = −iω0 − (ρgg − ρee ) . ρeg + i dt 2 h ¯ These equations are often called optical Bloch equations. At steady-state, ρee and ρgg tend to a constant value, while ρeg and ρge oscillate respectively as e−iωL t and eiωL t . This steady-state is reached after a characteristic time of the order of Γ −1 . ¿From the second equation we extract the steady-state value of ρeg as a function of ρgg − ρee = 1 − 2ρee : ρeg = i

h d · E(r) e−iωL t /¯ (1 − 2ρee ) . i∆ + Γ/2

We now insert this value in the evolution of ρee and we get:

272

26 Laser Cooling and Trapping

ρee =

s 2(1 + s)

with

s(r) =

h2 2 |d · E(r)|2 /¯ . ∆2 + Γ 2 /4

The three other values given in the text for ρgg , ρeg and ρge follow immediately. 26.1.4. The steady state value of ρee gives the average probability of ﬁnding the atom in the internal state |e. This value results from the competition between absorption processes, which tend to populate the level |e and stimulated+spontaneous emission processes, which depopulate |e to the beneﬁt of |g. The quantity Γ ρee represents the steady-state rate of spontaneous emission as the atom is irradiated by the laser wave. For a low saturation parameter s, this rate is proportional to the laser intensity |E(r)|2 . When the laser intensity increases, s gets much larger than 1 and the steady state value of ρee is close to 1/2. This means that the atom spends half of the time in level |e. In this case, the rate of spontaneous emission tends to Γ/2. Section 26.2: The Radiation Pressure Force 26.2.1. For a plane laser wave the force operator is given by: ˆ (r) = ik d · E 0 ei(k·r−ωL t) |eg| − e−i(k·r−ωL t) |ge| . F ˆ ) which gives: The expectation value in steady state is Tr(ˆ ρF f = F = ik d · E 0 ei(k·r−ωL t) ρge + c.c. s0 Γ =h ¯k 2 1 + s0 with s0 =

h2 2 |d · E 0 |2 /¯ . 2 2 ∆ + Γ /4

26.2.2. The interpretation of this result is as follows. The atom undergoes absorption processes, where it goes from the ground internal state to the excited internal state, and gains the momentum h ¯ k. From the excited state, it can return to the ground state by a stimulated or spontaneous emission process. In a stimulated emission the atom releases the momentum that it has gained during the absorption process, so that the net variation of momentum in a such a cycle is zero. In contrast, in a spontaneous emission process, the momentum change of the atom has a random direction and it averages to zero since the spontaneous emission process occurs with the same probability in two opposite directions. Therefore the net momentum gain for the atom in a cycle “absorption–spontaneous emission” is ¯hk corresponding to a velocity change vrec . Since these cycles occur with a rate (Γ/2)s0 /(1 + s0 ) (as found at the end of Sect. 26.1), we recover the expression for the radiation force found above.

26.5 Solutions

273

26.2.3. For a large laser intensity, the force saturates to the value ¯hkΓ/2. ¯ kΓ/(2m) = 9 × 105 m s−2 , which This corresponds to an acceleration amax = h is 100 000 times larger than the acceleration due to gravity. 26.2.4. In the rest frame of the atom, the laser ﬁeld still corresponds to a plane wave with a modiﬁed frequency ωL − k · v (ﬁrst order Doppler eﬀect). The change of momentum of the photon is negligible for non-relativistic atomic velocities. The previous result is then changed into: f =h ¯k

s(v) Γ 2 1 + s(v)

with

s(v) =

2 |d · E 0 |2 /¯ h2 . (∆ − k · v)2 + Γ 2 /4

26.2.5. The notion of force derived above is valid if the elementary velocity change in a single absorption or emission process (the recoil velocity vrec = h ¯ k/m) modiﬁes only weakly the value of f . This is the case when ¯ k 2 /m is very small compared the elementary change of Doppler shift kvrec = h with the width of the resonance: ¯ k2 h Γ . m This is the so called broad line condition. This condition is well satisﬁed for sodium atoms since ¯hk 2 /(mΓ ) = 5 × 10−3 in this case. Section 26.3: Doppler Cooling 26.3.1. The total force acting on the atom moving with velocity v is

|d · E 0 |2 /¯ |d · E 0 |2 /¯ h2 h2 f (v) = h ¯k Γ − , (∆ − kv)2 + Γ 2 /4 (∆ + kv)2 + Γ 2 /4 where we have used the fact that s 1. For low velocities (kv Γ ) we get at ﬁrst order in v f (v) = −

mv τ

with

τ=

m ∆2 + Γ 2 /4 . hk 2 s0 2(−∆)Γ ¯

This corresponds to a damping force if the detuning ∆ is negative. In this case the atom is cooled because of the Doppler eﬀect. This is the so-called Doppler cooling: A moving atom feels a stronger radiation pressure force from the counterpropagating wave than from the copropagating wave. For an atom at rest the two radiation pressure forces are equal and opposite: the net force is zero. 26.3.2. For a ﬁxed saturation parameter s0 , the cooling time is minimal for ∆ = −Γ/2, which leads to τmin =

m . 2¯ hk 2 s0

274

26 Laser Cooling and Trapping

Note that this time is always much longer than the lifetime of the excited state Γ −1 when the broad line condition is fullﬁlled. For sodium atoms this minimal cooling time is 16 µs for s0 = 0.1. 26.3.3. (a) The probability that an atom moving with velocity v scatters a photon from the ±z wave during the time dt is Γ s0 dP± (v) = 2

2∆kv 1± 2 ∆ + Γ 2 /4

dt .

Since we assume that the spontaneously emitted photons also propagate along z, half of the scattering events do not change the velocity of the atom: This is the case when the spontaneously emitted photon propagates along the same direction as the absorbed photon. For the other half of the events, the change of the atomic velocity is ±2vrec , corresponding to a spontaneously emitted photon propagating in the direction opposite to the absorbed photon. Consequently, the probability that the velocity of the atom does not change during the time dt is 1 − (dP+ (v) + dP− (v))/2, and the probability that the atomic velocity changes by ±2vrec is dP± (v)/2. Therefore one has:

P (v, t + dt) =

dP+ (v) + dP− (v) P (v, t) 2 dP+ (v − 2vrec ) P (v − 2vrec , t) + 2 dP− (v + 2vrec ) P (v + 2vrec , t) . + 2 1−

(b) Assuming that P (v) varies smoothly over the recoil velocity scale (which will be checked in the end), we can transform the ﬁnite diﬀerence equation found above into a diﬀerential equation: ∂ ∂2P ∂P = α (vP ) + β 2 , ∂t ∂v ∂v with m 2 β = Γ vrec s0 . τ The term proportional to α corresponds to the Doppler cooling. The term in β accounts for the heating due to the random nature of spontaneous emission processes. The coeﬃcient β is a diﬀusion constant in velocity space, proportional to the square of the elementary step of the random walk vrec , and to its rate Γ s0 . (c) The steady state for P (v) corresponds to the solution of: α=

26.5 Solutions

275

dP =0, dv whose solution (for α > 0, i.e. ∆ < 0) is a Maxwell distribution:

αv 2 P (v) = P0 exp − . 2β α v P (v) + β

The eﬀective temperature is therefore kB T =

h ∆2 + Γ 2 /4 ¯ mβ = . α 2 −∆

(d) The minimal temperature is obtained for ∆ = −Γ/2: ¯Γ h . 2 This is the Doppler cooling limit, which is independent of the laser intensity. Note that, when the broad line condition is fullﬁlled, the corresponding velocity scale v0 is such that: hΓ/m Γ/k . vrec v0 = ¯ kB Tmin =

The two hypotheses at the basis of our calculation are therefore valid: (i) P (v) varies smootly over the scale vrec and (ii) the relevant velocities are small enough for the linearization of the scattering rates to be possible. For sodium atoms, the minimal temperature is Tmin = 240 µK, corresponding to v0 = 40 cm s−1 . Section 26.4: The Dipole Force 26.4.1. For a real amplitude E 0 (r) of the electric ﬁeld of the light wave ˆ (r) is: (standing wave), the force operator F ⎞ ⎛

ˆ (r) = ⎝ di ∇E0i (r)⎠ e−iωL t |eg| + eiωL t |ge| . F i=x,y,z

Assuming that the internal dynamics of the atom has reached its steady-state ˆ: value, we get for the expectation value of F f (r) = F = −∇(d · E 0 (r)) =−

∆ d · E 0 (r) 2 1 + s(r) ∆ + Γ 2 /4

¯ ∆ ∇s(r) h 2 1 + s(r)

26.4.2. This force is called the dipole force. It derives from the dipole potential U (r):

276

26 Laser Cooling and Trapping

f (r) = −∇U (r)

with

U (r) =

¯∆ h log(1 + s(r)) . 2

For a laser ﬁeld with intensity P = 1 W, focused on a spot with radius r = 10 µm, the electric ﬁeld at the center is 2P E0 = = 1.6 × 106 V/m . π0 cr2 We suppose here that the circular spot is uniformly illuminated. A more accurate treatment should take into account the transverse Gaussian proﬁle of the laser beam, but this would not signiﬁcantly change the following reh = 3.1 × 1011 s−1 and the detuning ∆ is sults. This value for E0 leads to dE0 /¯ 14 −1 equal to 3×10 s . The potential depth is then found to be equal to 2.4 mK, 10 times larger than the Doppler cooling limit. Due to the large detuning, the photon scattering rate is quite small: 70 photons/s.

26.6 Comments The radiation pressure force has been used in particular for atomic beam deceleration (J.V. Prodan, W.D. Phillips, and H. Metcalf, Phys. Rev. Lett. 49, 1149 (1982)). The Doppler cooling was proposed by T.W. H¨ ansch and A. Schawlow (Opt. Commun. 13, 68 (1975)). A related cooling scheme for trapped ions was proposed the same year by D. Wineland and H. Dehmelt (Bull. Am. Phys. Soc. 20, 637 (1975)). The ﬁrst observation of 3D laser cooling of neutral atoms was reported by S. Chu, L. Hollberg, J.E. Bjorkholm, A. Cable, and A. Ashkin, Phys. Rev. Lett. 55, 48 (1985), and the same group reported one year later the observation of atoms trapped at the focal point of a laser beam using the dipole force (Phys. Rev. Lett. 57, 314 (1986)). It was subsequently discovered experimentally in the group of W.D. Phillips that the temperature of laser cooled atoms could be much lower than ¯ Γ/2 derived in this problem. This clear violation the Doppler limit kB T = h of Murphy’s law (an experiment working 10 times better than predicted!) was explained independently in terms of Sisyphus cooling by the groups of C. Cohen-Tannoudji and S. Chu (for a review, see e.g. C. Cohen-Tannoudji and W.D. Phillips, Physics Today, October 1990, p.33). The Physics Nobel Prize was awarded in 1997 to S. Chu, C. CohenTannoudji and W.D. Phillips for their work on the trapping and cooling of atoms with laser light.

27 Bloch Oscillations

The possibility to study accurately the quantum motion of atoms in standing light ﬁelds has been used recently in order to test several predictions relating to wave propagation in a periodic potential. We present in this chapter some of these observations related to the phenomenon of Bloch oscillations.

27.1 Unitary Transformation on a Quantum System Consider a system in the state |ψ(t) which evolves under the eﬀect of a ˆ ˆ Hamiltonian H(t). Consider a unitary operator D(t). Show that the evolution of the transformed vector ˜ ˆ |ψ(t) = D(t)|ψ(t) is given by a Schr¨ odinger equation with Hamiltonian ˆ dD(t) ˜ ˆ † (t) . ˆ ˆ H(t) ˆ D ˆ † (t) + i¯ D H(t) = D(t) h dt

27.2 Band Structure in a Periodic Potential The mechanical action of a standing light wave onto an atom can be described by a potential (see e.g. Chap. 26). If the detuning between the light frequency and the atom resonance frequency ωA is large compared to the electric dipole coupling of the atom with the wave, this potential is proportional to the light intensity. Consequently, the one-dimensional motion of an atom of mass m moving in a standing laser wave can be written ˆ2 ˆ , ˆ = P + U0 sin2 (k0 X) H 2m

278

27 Bloch Oscillations

ˆ and Pˆ are the atomic position and momentum operators and where where X we neglect any spontaneous emission process. We shall assume that k0 ωA /c ¯ 2 k02 /(2m). and we introduce the “recoil energy” ER = h ˆ recall brieﬂy why 27.2.1. (a) Given the periodicity of the Hamiltonian H, the eigenstates of this Hamiltonian can be cast in the form (Bloch theorem): |ψ = eiqX |uq , ˆ

where the real number q (Bloch index) is in the interval (−k0 , k0 ) and where |uq is periodic in space with period λ0 /2. (b) Write the eigenvalue equation to be satisﬁed by |uq . Discuss the corresponding spectrum (i) for a given value of q, (ii) when q varies between −k0 and k0 . ˆ are denoted |n, q, with energies En (q). In the following, the eigenstates of H They are normalized on a spatial period of extension λ0 /2 = π/k0 . 27.2.2. Give the energy levels in terms of the indices n and q in the case U0 = 0. 27.2.3. Treat the eﬀect of the potential U0 in ﬁrst order perturbation theory, for the lowest band n = 0 (one should separate the cases q = ±k0 and q “far from” ±k0 ). Give the width of the gap which appears between the bands n = 0 and n = 1 owing to the presence of the perturbation. 27.2.4. Under what condition on U0 is this perturbative approach reliable? 27.2.5. How do the widths of the other gaps vary with U0 in this perturbative limit?

27.3 The Phenomenon of Bloch Oscillations We suppose now that we prepare in the potential U0 sin2 (k0 x) a wave packet in the n = 0 band with a sharp distribution in q, and that we apply to the atom a constant extra force F = ma. We recall the adiabatic theorem: suppose that a system is prepared at time (0) ˆ ˆ If the Hamiltonian H(t) 0 in the eigenstate |φn of the Hamiltonian H(0). evolves slowly with time, the system will remain with a large probability in (t) (t) (t) ¯ φm |φ˙ n the eigenstate |φn . The validity condition for this theorem is h (t) (t) d |Em (t) − En (t)| for any m = n. We use the notation |φ˙ n = dt |φn . 27.3.1. Preparation of the Initial State. Initially U0 = 0, a = 0 and the atomic momentum distribution has a zero average and a dispersion small

27.3 The Phenomenon of Bloch Oscillations

279

compared to ¯hk. We will approximate this state by the eigenstate of momentum |p = 0. One “slowly” switches on the potential U0 (t) sin2 (k0 x), with U0 (t) ≤ ER . (a) Using the symmetries of the problem, show that the Bloch index q is a constant of the motion. (b) Write the expression of the eigenstate of H(t) of indices n = 0, q = 0 to ﬁrst order in U0 . (c) Evaluate the validity of the adiabatic approximation in terms of U˙ 0 , ER , ¯h. (d) One switches on linearly the potential U0 until it reaches the value ER . What is the condition on the time τ of the operation in order for the process to remain adiabatic? Calculate the minimal value of τ for cesium atoms (m = 2.2 × 10−25 kg, λ0 = 0.85 µm.) 27.3.2. Devising a Constant Force. Once U0 (t) has reached the maximal value U0 (time t = 0), one achieves a sweep of the phases φ+ (t) and φ− (t) of the two traveling waves forming the standing wave. The potential seen by the atom is then U0 sin2 (k0 x − (φ+ (t) − φ− (t))/2) and one chooses φ+ (t) − φ− (t) = k0 at2 . (a) Show that there exists a reference frame where the wave is stationnary, and give its acceleration. (b) In order to study the quantum motion of the atoms in the accelerated reference frame, we consider the unitary transformation generated by ˆ h) exp(ima2 t3 /(3¯ ˆ h) exp(−imatX/¯ D(t) = exp(iat2 Pˆ /2¯ h)) . ˆ and Pˆ transform? Write How do the position and momentum operators X the resulting form of the Hamiltonian in this unitary transformation ˆ2 ˜ ˆ = P + U0 sin2 (k0 X) ˆ + maX ˆ . H 2m

27.3.3. Bloch Oscillations We consider the evolution of the initial state n = 0, q = 0 under the eﬀect of ˜ ˆ the Hamiltonian H. (a) Check that the state vector remains of the Bloch form, i.e. |ψ(t) = eiq(t)X |u(t) , ˆ

where |u(t) is periodic in space and q(t) = −mat/¯h. (b) What does the adiabatic approximation correspond to for the evolution of |u(t)? We shall assume this approximation to be valid in the following.

280

27 Bloch Oscillations

(c) Show that, up to a phase factor, |ψ(t) is a periodic function of time, and give the corresponding value of the period. (d) The velocity distribution of the atoms as a function of time is given in Fig. 27.1. The time interval between two curves is 1 ms and a = −0.85 ms−2 . Comment on this ﬁgure, which has been obtained with cesium atoms.

Fig. 27.1. Atomic momentum distribution of the atoms (measured in the accelerated reference frame) for U0 = 1.4 ER . The lower curve corresponds to the end of the preparation phase (t = 0) and the successive curves, from bottom to top, are separated by time intervals of one millisecond. For clarity, we put a diﬀerent vertical oﬀset for each curve

27.4 Solutions

281

27.4 Solutions Section 27.1: Unitary Transformation on a Quantum System ˜ gives The time derivative of |ψ ˙ = i¯ ˜˙ = i¯ ˆ˙ D ˆ† + D ˆ˙ ˆ ψ ˆ ˆH ˆD ˆ † D|ψ + D| h D h D|ψ , i¯ h|ψ hence the results of the lemma. Section 27.2: Band Structure in a Periodic Potential 27.2.1. Bloch theorem (a) The atom moves in a spatially periodic potential, with period λ0 /2 = π/k0 . Therefore the Hamiltonian commutes with the translation operator h)) and one can look for a common basis set of Tˆ(λ0 /2) = exp(iλ0 Pˆ /(2¯ these two operators. Eigenvalues of Tˆ(λ0 /2) have a modulus equal to 1, since Tˆ(λ0 /2) is unitary. They can be written eiqλ0 /2 where q is in the interval ˆ and Tˆ(λ0 /2) is then such that (−k0 , k0 ). A corresponding eigenvector of H Tˆ(λ0 /2)|ψ = eiqλ0 /2 |ψ or in other words ψ(x + λ0 /2) = eiqλ0 /2 ψ(x) . This amounts to saying that the function uq (x) = e−iqx ψ(x) is periodic in space with period λ0 /2, hence the result. (b) The equation satisﬁed by uq is: −

¯2 h 2m

d + iq dx

2 uq + U0 sin2 (kx) uq = E uq .

For a ﬁxed value of q, we look for periodic solutions of this equation. The boundary conditions uq (λ0 /2) = uq (0) and uq (λ0 /2) = uq (0) lead, for each q, to a discrete set of allowed values for E, which we denote En (q). The ˆ and Tˆ(λ0 /2) is denoted |ψ = |n, q. Now corresponding eigenvector of H when q varies in the interval (−k0 , k0 ), the energy En (q) varies continuously in an interval (Enmin , Enmax ). The precise values of Enmin and Enmax depend on the value of U0 . The spectrum En (q) is then constituted by a series of allowed energy bands, separated by gaps corresponding to forbidden values of energy. The interval (−k0 , k0 ) is called the ﬁrst Brillouin zone. ˆ is simply ¯h2 k 2 /(2m) corresponding 27.2.2. For U0 = 0, the spectrum of H ikx (free particle). Each k can be written: k = q + 2nk0 to the eigenstates e where n is an integer, and the spectrum En (q) then consists of folded portions of parabola (see Fig. 27.2a). There are no forbidden gaps in this case, and the min = Enmax ). various energy bands touch each other (En+1

282

27 Bloch Oscillations

Fig. 27.2. Structure of the energy levels En (q) (a) for U0 = 0 and (b) U0 = ER

ˆ has no degeneracy, 27.2.3. When q is far enough from ±k0 , the spectrum of H and the shift of the energy level of the lowest band n = 0 can be obtained using simply: ˆ ∆E0 (q) = 0, q|U0 sin2 (k0 X)|0, q π/k0 U0 k0 . e−iqx U0 sin2 (k0 x) eiqx dx = = π 0 2 When q is equal to ±k0 , the bands n = 0 and n = 1 coincide and one should diagonalize the restriction of U (x) to this two-dimensional subspace. One gets U0 ˆ ˆ k0 = 1, k0 |U0 sin2 (k0 X)|1, k0 = , 0, k0 |U0 sin2 (k0 X)|0, 2 U0 ˆ ˆ 0, k0 |U0 sin2 (k0 X)|1, k0 = 1, k0 |U0 sin2 (k0 X)|0, k0 = − . 4 The diagonalization of the matrix

U0 2 −1 4 −1 2 gives the two eigenvalues 3U0 /4 and U0 /4, which means that the two bands n = 0 and n = 1 do not touch each other anymore, but that they are separated by a gap U0 /2 (see Fig. 27.2b for U0 = ER ). 27.2.4. This perturbative approach is valid if one can neglect the coupling to all other bands. Since the characteristic energy splitting between the band n = 1 and the band n = 2 is 4 ER , the validity criterion is U0 4 ER . 27.2.5. The other gaps open either at k = 0 or k = ±k0 . They result from the coupling of eink0 x and e−ink0 x under the inﬂuence of U0 sin2 (k0 x). This coupling gives a non-zero result when taken at order n. Therefore the other gaps scale as U0n and they are much smaller that the lowest one.

27.4 Solutions

283

Section 27.3: The phenomenon of Bloch Oscillations 27.3.1. Preparation of the Initial State (a) Suppose that the initial state has a well deﬁned Bloch index q, which means that Tˆ(λ0 /2)|ψ(0) = eiqλ0 /2 |ψ(0) . ˆ At any time t, the Hamiltonian H(t) is spatially periodic and commute with ˆ (t) also the translation operator Tˆ(λ0 /2). Therefore the evolution operator U commutes with Tˆ(λ0 /2). Consequently: ˆ (t)|ψ(0) = U ˆ (t)Tˆ(λ0 /2)|ψ(0) , Tˆ(λ0 /2)|ψ(t) = Tˆ(λ0 /2)U = eiqλ0 /2 |ψ(t) , which means that q is a constant of motion. (b) At zeroth order in U0 , the eigenstates of H corresponding to the Bloch index q = 0 are the plane waves |k = 0 (energy 0), |k = ±2k0 (energy 4ER ), . . . At ﬁrst order in U0 , in order to determine |n = 0, q = 0, we have to take into account the coupling of |k = 0 with |k = ±2k0 , which gives |n = 0, q = 0 = |k = 0 +

k = 2k0 |U0 sin2 k0 x|k = 0 |k = 2k0 . 4ER =±

The calculation of the matrix elements is straightforward and it leads to x|n = 0, q = 0 ∝ 1 +

U0 (t) cos(2k0 x) . 8ER

(c) The system will adiabatically follow the level |n = 0, q = 0 as the potential U0 is raised, provided for any n

d|n = 0, q = 0 hn , q = 0| ¯ En (0) − E0 (0) . dt Using the value found above for |n = 0, q = 0 and taking n = ±1, we derive the validity criterion for the adiabatic approximation in this particular case: 2 . hU˙ 0 64 ER ¯

(d) For a linear variation of U0 such that U0 = ER t/τ , this validity condition is τ ¯ h/(64 ER ) , which corresponds to τ 10 µs for cesium atoms.

284

27 Bloch Oscillations

27.3.2. Devising a Constant Force (a) Consider a point with coordinate x in the lab frame. In the frame with acceleration a and zero initial velocity, the coordinate of this point is x = x − at2 /2. In this frame, the laser intensity varies as sin2 (kx ), corresponding to a “true” standing wave. ˆ f (Pˆ )] = i¯ ˆ = (b) Using the standard relations [X, hf (Pˆ ) and [Pˆ , g(X)] ˆ one gets −i¯ hg (X), 2 ˆ Pˆ D ˆ † = Pˆ + mat . ˆX ˆD ˆ† = X ˆ + at D D 2 ˆH ˆD ˆ † is The transformed Hamiltonian D

2 ˆ , ˆH ˆD ˆ † = 1 Pˆ + mat + U0 sin2 (k0 X) D 2m ˜ ˆ can be written and the extra term appearing in H 2 2 ˆ dD ˆ † = −atPˆ + maX ˆ − ma t . D dt 2 Summing the two contributions, we obtain

i¯ h

ˆ2 ˜ ˆ = P + U0 sin2 (k0 X) ˆ + maX ˆ . H 2m This Hamiltonian describes the motion of a particle of mass m in a periodic potential, superimposed with a constant force – ma. 27.3.3. Bloch Oscillations ˜ˆ ˙ = H|ψ. (a) The evolution of the state vector is i¯ h|ψ We now put |ψ(t) = ˆ h)|u(t) and we look for the evolution of |u(t). We obtain after exp(−imatX/¯ a straightforward calculation h2 ¯ ∂u(x, t) =− i¯ h ∂t 2m

imat ∂ − ∂x h ¯

2 u(x, t) + U0 sin2 (k0 x) u(x, t) .

Using the structure of this equation, and using the initial spatial periodicity of u(x, 0), one deduces that u(x, t) is also spatially periodic with the same period λ0 /2. (b) The adiabatic hypothesis for |u(t) amounts to assume that this vector, which is equal to |un=0,q=0 at t = 0, remains equal to |u0,q(t) at any time. The atom stays in the band n = 0. hk0 /(ma) during which q(t) is changed into (c) Consider the duration TB = 2¯ q(t) − 2k0 . Since 2k0 is the width of the Brillouin zone, we have |un,q−2k0 ≡ |un,q . Consequently, when the adiabatic approximation is valid, the state |ψ(t + TB ) coincides (within a phase factor) with the state |ψ(t). Since this phase factor does not enter in the calculation of physical quantities such as

27.5 Comments

285

position or momentum distributions, we expect that the evolution of these quantities with time will be periodic with the period TB . (d) We ﬁrst note that the initial distribution is such that the average momentum is zero, and that the momentum dispersion is small compared with hk0 , as assumed in this problem. Concerning the time evolution, we see in¯ deed that the atomic momentum distribution is periodic in time, with a period TB 8 ms, which coincides with the predicted value 2¯hk0 /(ma). Finally we note that the average momentum increases quasi-linearly with time during the ﬁrst 4 ms, from 0 to ¯hk0 . At this time corresponding to TB /2, a “reﬂexion” occurs and the momentum is changed into −¯ hk0 . During the second half of the Bloch period (from 4 ms to 8 ms) the momentum again increases linearly with time from −¯ hk0 to 0. At the time TB /2, the particle is at the edge of the Brillouin zone (q = ±k0 ). This is the place where the adiabatic approximation is the most fragile since the band n = 1 is then very close to the band n = 0 (gap U0 ). One can check that the validity criterion for the adiabatic approximation at this place is maER k0 U02 , which is well fullﬁlled in the experiment. The reﬂection occurring at t = TB /2 can be viewed as a Bragg reﬂection of the atom with momentum h ¯ k0 on the periodic grating U0 sin2 (kx).

27.5 Comments This paradoxical situation, where a constant force ma leads to an oscillation of the particle instead of a constant acceleration, is called the Bloch oscillation phenomenon. It shows that an ideal crystal cannot be a good conductor: when one applies a potential diﬀerence at the edge of the crystal, the electrons of the conduction band feel a constant force in addition to the periodic potential created by the crystal and they should oscillate instead of being accelerated towards the positive edge of the crystal. The conduction phenomenon results from the defects present in real metals. The experimental data have been extracted from M. Ben Dahan, E. Peik, J. Reichel, Y. Castin, and C. Salomon, Phys. Rev. Lett. 76, 4508 (1996) and from E. Peik, M. Ben Dahan, I. Bouchoule, Y. Castin, and C. Salomon, Phys. Rev. A 55, 2989 (1997). A review of atom optics experiments performed with standing light waves is given in M. Raizen, C. Salomon, and Q. Niu, Physics Today, July 1997, p. 30.

1 Neutrino Oscillations

In β decay or, more generally, in Weak interactions, the electron is always associated with a neutral particle, the neutrino νe . There exists in nature another particle, the µ lepton, or muon, whose physical properties seem completely analogous to those of the electron, except for its mass mµ 200 me . The muon has the same Weak interactions as the electron, but it is associated to a diﬀerent neutrino, the νµ . A neutrino beam produced in an accelerator can interact with a neutron (n) in a nucleus and give rise to the reactions νe + n → p + e

and

νµ + n → p + µ ,

(1.1)

whereas the reactions νe + n → p + µ or νµ + n → p + e are never observed. The reactions (1.1) are used in practice in order to detect neutrinos. Similarly, a π − meson can decay via the modes π − → µ + ν¯µ

(dominant mode) and π − → e + ν¯e ,

(1.2)

whereas π − → µ + ν¯e or π − → e + ν¯µ are never observed. This is how one can produce neutrinos abundantly (it is easy to produce π mesons). In (1.2) we have introduced the antiparticles ν¯µ et ν¯e . There is a (quasi) strict symmetry between particles and their antiparticles, so that, in the same way as the electron is associated with the neutrino νe , the antielectron, or positron, e+ is associated with the antineutrino ν¯e . One observes the “charge-conjugate” reactions of (1.1) and (1.2) ν¯e + p → n + e+ ,

ν¯µ + p → n + µ+

and π + → µ+ + νµ .

(1.3)

In all what follows, what we will say about neutrinos holds symmetrically for antineutrinos. In 1975, a third lepton, the τ , was discovered. It is much more massive, mµ 3500 me , it is associated with its own neutrino ντ , and it obeys the same physical laws as the two lighter leptons, except for mass eﬀects. Since the

18

1 Neutrino Oscillations

1990’s, the experimental measurements at the LEP colliding ring in CERN have shown that these three neutrinos νe , νµ , ντ (and their antiparticles) are the only ones of their kinds (at least for masses less that 100 GeV/c2 ). For a long time, physicists believed that neutrinos were zero-mass particles, as is the photon. In any case, their masses (multiplied by c2 ) are considerably smaller than the energies involved in experiments where they are observed. Therefore, many experimental limits on these masses are consistent with zero. However, both theoretical and cosmological arguments suggested that this might not be the case. The proof that neutrino masses are not all zero is a great discovery of the last ten years. In the present study, we show how the mass diﬀerences of neutrinos can be measured by a quantum oscillation eﬀect. The idea is that the “ﬂavor” neutrinos νe , νµ and ντ , which are produced or detected experimentally are not eigenstates of the mass, but rather linear combinations of mass eigenstates ν1 , ν2 , ν3 , with masses m1 , m2 , m3 . The neutrinos observed on earth have various origins. They can be produced in accelerators, in nuclear reactors, and also in the atmosphere by cosmic rays, or in thermonuclear reaction inside stars, in particular the core of the sun, and in supernovae explosions.

1.1 Mechanism of the Oscillations; Reactor Neutrinos In this ﬁrst part, we consider oscillations between two types of neutrinos, the νe and the νµ . This simple case will allow us to understand the underlying physics of the general case. We will analyze the data obtained with nuclear reactors. The average energy of the (anti-)neutrinos produced in reactors is E = 4 MeV, with a dispersion of the same order. In all what follows, we will assume that if m is the neutrino mass and p and E its momentum and energy, the mass is so small that the energy of a neutrino of mass m and momentum p is E=

p2 c2 + m2 c4 pc +

m2 c4 , 2pc

(1.4)

and that the neutrino propagates to very good approximation at the velocity of light c. ˆ be the Hamiltonian of a free neutrino of momentum p, which we Let H ˆ assume to be well deﬁned. We note |ν1 and |ν2 the two eigenstates of H: ˆ j = Ej |νj , H|ν

Ej = pc +

m2j c4 , 2pc

j = 1, 2 .

m1 and m2 are the respective masses of the states |ν1 and |ν2 , and we assume m1 = m2 .

1.1 Mechanism of the Oscillations; Reactor Neutrinos

19

The oscillations of freely propagating neutrinos come from the following quantum eﬀect. If the physical states of the neutrinos which are produced (reactions (1.2)) or detected (reactions (1.1)) are not |ν1 and |ν2 , but linear combinations of these: |νe = |ν1 cos θ + |ν2 sin θ ,

|νµ = −|ν1 sin θ + |ν2 cos θ

(1.5)

where θ is a mixing angle to be determined, these linear combination of energy eigenstates oscillate in time and this leads to measurable eﬀects. 1.1.1. At time t = 0, one produces a neutrino of momentum p in the state |νe . Calculate the state |ν(t) at time t in terms of |ν1 and |ν2 . 1.1.2. What is the probability Pe for this neutrino to be detected in the state |νe at time t? The result will be expressed in terms of the mixing angle θ and of the oscillation length L L=

4π¯ hp , |∆m2 | c2

∆m2 = m21 − m22 .

(1.6)

1.1.3. Calculate the oscillation length L for an energy E pc = 4 MeV and a mass diﬀerence ∆m2 c4 = 10−4 eV2 . 1.1.4. One measures the neutrino ﬂuxes with a detector located at a distance from the production area. Express the probability Pe as a function of the distance = ct. 1.1.5. The mass of the muon satisﬁes mµ c2 = 106 MeV. Conclude that in such an experiment one cannot detect muon neutrinos νµ with the reaction (1.1). We recall that mp c2 = 938.27 MeV and mn c2 = 939.57 MeV. 1.1.6. The detectors measure neutrino ﬂuxes with an accuracy of ∼ 10%. (a) Assuming ∆m2 c4 = 10−4 eV2 , determine the minimal distance min where to put a detector in order to detect an oscillation eﬀect. For this calculation, assume the mixing in (1.5) is maximum, i.e. θ = π/4. (b) How does min change if the mixing is not maximum? 1.1.7. Several experiments on neutrinos produced by nuclear energy plants have been performed in Chooz and in Bugey in France. The most recent data comes from the KamLAND collaboration, in Japan. The results are given on Fig. 1.1. (a) Explain the results of Fig. 1.1, except that of KamLAND. (b) The KamLAND experiment, which was performed in 2002, consisted in measuring the neutrinos coming from all the (numerous) reactors in Japan and neighboring countries, which amounts to taking an average distance of = 180 km. Putting together that data and the results of numerous experiments performed on solar neutrinos, the physicists of Kamland come to the following results:

20

1 Neutrino Oscillations

Ndetected N expected

1,2 1,0 0,8 0,6

ILL Bugey Rovno Goesgen Chooz KamLAND

0,4 0,2 0 10

102

distance (meters) 103

104

105

Fig. 1.1. Ratio between the numbers of observed electron neutrinos and those expected in the absence of oscillations as a function of the distance to the reactor

|∆m2 | c4 = 7.1 (± 0.4) × 10−5 eV2 ,

tan2 θ = 0.45 (± 0.02) . (1.7)

Show that these values are consistent with the result Pe = 0.61 (± 0.10) of Fig. 1.1.

1.2 Oscillations of Three Species; Atmospheric Neutrinos We now consider the general formalism with three neutrino species. We denote |να , α = e, µ, τ the “ﬂavor” neutrinos and |νi , i = 1, 2, 3 the mass eigenstates. These two bases are related to one another by the Maki-Nagawakaˆ, Sakata (MNS) matrix U ⎞ ⎛ Ue1 Ue2 Ue3 3 ⎟ ˆ =⎜ Uαi |νi , U |να = (1.8) ⎝ Uµ1 Uµ2 Uµ3 ⎠ i=1 Uτ 1 Uτ 2 Uτ 3 ∗ This matrix is unitary ( i Uβi Uαi = δαβ ) and it can be written as: ⎞ ⎞⎛ ⎞⎛ ⎛ 0 s13 e−iδ c13 c12 s12 0 1 0 0 ˆ = ⎝ 0 c23 s23 ⎠ ⎝ ⎠ ⎝ −s12 c12 0 ⎠ 0 1 0 U iδ 0 −s23 c23 0 0 1 −s13 e 0 c13 where cij = cos θij and sij = sin θij . The complete experimental solution of the problem would consist in measuring the three mixing angles θ12 , θ23 , θ13 , the phase δ, and the three masses m1 , m2 , m3 . We consider situations such that (1.4) is still valid. 1.2.1. At time t = 0 a neutrino is produced with momentum p in the state |ν(0) = |να . Express, in terms of the matrix elements Uαi , its state at a

1.2 Oscillations of Three Species; Atmospheric Neutrinos

21

later time t. Write the probability Pα→β (t) to observe a neutrino of ﬂavor β at time t. 1.2.2. We deﬁne the oscillation lengths at an energy E pc by: Lij =

4π¯ hp , |∆m2ij |c2

∆m2ij = m2i − m2j .

(1.9)

Notice that there are only two independent oscillation lengths since ∆m212 + ∆m223 + ∆m231 = 0. For neutrinos of energy E = 4 GeV, calculate the oscillation lengths L12 and L23 . We will choose for |∆m212 | the result given in (1.7), and we will choose |∆m223 | c4 = 2.5 × 10−3 eV2 , a value which will be justiﬁed later on. 1.2.3. The neutrino counters have an accuracy of the order of 10% and the energy is E = 4 GeV. Above which distances 12 and 23 of the production point of the neutrinos can one hope to detect oscillations coming from the superpositions 1 ↔ 2 and 2 ↔ 3? 1.2.4. The Super-Kamiokande experiment, performed in 1998, consists in detecting “atmospheric” neutrinos. Such neutrinos are produced in the collision of high energy cosmic rays with nuclei in the atmosphere at high altitudes. In a series of reactions, π ± mesons are produced abundantly, and they decay through the chain: π − → µ− + ν¯µ

followed by

µ− → e− + ν¯e + νµ ,

(1.10)

and an analogous chain for π + mesons. The neutrino ﬂuxes are detected in an underground detector by the reactions (1.1) and (1.3). To simplify things, we assume that all muons decay before reaching the surface of the Earth. Deduce that, in the absence of neutrino oscillations, the expected ratio between electron and muon neutrinos Rµ/e =

νµ ) N (νµ ) + N (¯ N (νe ) + N (¯ νe )

would be equal to 2. 1.2.5. The corrections to the ratio Rµ/e due to the fact that part of the muons reach the ground can be calculated accurately. Once this correction is made, one ﬁnds, by comparing the measured and calculated values for Rµ/e (Rµ/e )measured = 0.64 (± 0.05) . (Rµ/e )calculated In order to explain this relative decrease of the number of νµ ’s, one can think of oscillations of the types νµ νe and νµ ντ . The Super-Kamiokande experiment consists in varying the time of ﬂight of the neutrinos by measuring selectively the direction where they come from, as indicated on Fig. 1.2. The

22

1 Neutrino Oscillations

Electron neutrinos 120 cosmic ray

zenithal angle α

80 40

neutrino

cos α

Detector 0

Earth

-1

-0,5

0

0,5

1

Muon neutrinos

300

200 atmosphere 100 cos α 0

-1

-0,5

0

0,5

1

Fig. 1.2. Left: production of atmospheric neutrinos in collisions of cosmic rays with terrestrial atmospheric nuclei. The underground detector measures the ﬂux of electron and muon neutrinos as a function of the zenithal angle α. Right: number of atmospheric neutrinos detected in the Super-Kamiokande experiment as a function of the zenithal angle (this picture is drawn after K. Tanyaka, XXII Physics in Collisions Conference, Stanford 2002)

neutrinos coming from above (cos α ∼ 1) have traveled a distance equal to the atmospheric height plus the depth of the detector, while those coming from the bottom (cos α ∼ −1) have crossed the diameter of the Earth (13 400 km). Given the weakness of the interaction of neutrinos with matter, one can consider that the neutrinos propagate freely on a measurable distance between a few tens of km and 13 400 km. The neutrino energies are typically 4 GeV in this experiment. Can one observe a νe νµ oscillation of the type studied in the ﬁrst part? 1.2.6. The angular distributions of the νe and the νµ are represented on Fig. 1.2, together with the distributions one would observe in the absence of oscillations. Explain why this data is compatible with the fact that one observes a νµ ντ oscillation, no νe ντ oscillation, and no νe νµ oscillation.

1.3 Solutions

23

1.2.7. In view of the above results, we assume that there is only a twoneutrino oscillation phenomenon: νµ ντ in such an observation. We therefore use the same formalism as in the ﬁrst part, except that we change the names of particles. By comparing the muon neutrino ﬂux coming from above and from below, give an estimate of the mixing angle θ23 . In order to take into account the large energy dispersion of cosmic rays, and therefore of atmospheric neutrinos, we replace the oscillating factor sin2 (π/L23 ) by its mean value 1/2 if L23 . The complete results published by the Super-Kamiokande experiment are |∆m223 | c4 = 2.5 × 10−3 eV2 ,

θ23 = π/4 ,

θ13 = 0 .

Do they agree with the above considerations?

1.3 Solutions Section 1.1: Mechanism of the Oscillations: Reactor Neutrinos 1.1.1. Initially, the neutrino state is |ν(0) = |νe = |ν1 cos θ + |ν2 sin θ. Therefore, we have at time t |ν(t) = |ν1 cos θ e−iE1 t/¯h + |ν2 sin θ e−iE2 t/¯h . 1.1.2. The probability to ﬁnd this neutrino in the state |νe at time t is 2 Pe (t) = |νe |ν(t)|2 = cos2 θ e−iE1 t/¯h + sin2 θ e−iE2 t/¯h , which gives, after a simple calculation:

Pe (t) = 1 − sin2 (2θ) sin2

(E1 − E2 )t 2¯h

.

We have E1 − E2 = (m21 − m22 )c4 /(2pc). Deﬁning the oscillation length by L = 4π¯ hp/(|∆m2 | c2 ), we obtain

πct Pe (t) = 1 − sin2 (2θ) sin2 . L 1.1.3. For an energy E = pc = 4 MeV and a mass diﬀerence ∆m2 c4 = 10−4 eV2 , we obtain an oscillation length L = 100 km. 1.1.4. The time of ﬂight is t = /c. The probability Pe () is therefore

π 2 2 Pe () = 1 − sin (2θ) sin . (1.11) L

24

1 Neutrino Oscillations

1.1.5. A νµ energy of only 4 MeV is below the threshold of the reaction νµ + n → p + µ. Therefore this reaction does not occur with reactor neutrinos, and one cannot measure the νµ ﬂux. 1.1.6. In order to detect a signiﬁcant decrease in the neutrino ﬂux νe , we must have

π sin2 (2θ) sin2 > 0.1 . L (a) For the maximum mixing θ = π/4, i.e. sin2 (2θ) = 1, this implies π/L > 0.32 or > L/10. For E = 4 MeV and ∆m2 c4 = 10−4 eV2 , one ﬁnds > 10 km. The typical distances necessary to observe this phenomenon are of the order of a fraction of the oscillation length. (b) If the mixing is not maximum, one must operate at distances greater than L/10. Note that is the mixing angle is too small, (sin2 (2θ) < 0.1 i.e. θ < π/10), the oscillation amplitude is too weak to be detected, whatever the distance . In that case, one must improve the detection eﬃciency to obtain a positive conclusion. 1.1.7. (a) In all experiments except KamLAND, the distance is smaller than 1 km. Therefore, in all of these experiments |1 − Pe | ≤ 10−3 . The oscillation eﬀect is not detectable if the estimate |∆m2 | c4 ∼ 10−4 eV2 is correct. (b) For |∆m2 | c4 = 7.1 × 10−5 eV2 , tan2 θ = 0.45 and = 180 km, we obtain Pe = 0.50 which agrees with the measurement. The theoretical prediction taking into account the eﬀects due to the dispersion in energy is drawn on Fig. 1.3. We see incidentally how important it is to control error bars in such an experiment.

Ndetected N expected

1,2 1,0 0,8 0,6

ILL Bugey Rovno Goesgen Chooz KamLAND

0,4 0,2 0 10

102

distance (meters) 103

104

105

Fig. 1.3. Experimental points of Fig. 1.1 and the theoretical prediction of (1.11) (sinusoidal function damped by energy dispersion aﬀects). This curve is a best ﬁt of solar neutrino data. We notice that the KamLAND data point corresponds to the second oscillation of the curve

1.3 Solutions

25

Section 1.2: Oscillations of Three Species: Atmospheric Neutrinos 1.2.1. At time t = 0, we have: |ν(0) = |να =

Uαj |νj ,

j

and therefore at time t: |ν(t) = e−ipct/¯h

2 3

Uαj e−imj c

t/(2¯ hp)

|νj .

j

We conclude that the probability Pα→β to observe a neutrino of ﬂavor β at time t is 2 2 3 2 ∗ −imj c t/(2¯ hp) Uβj Uαj e Pα→β (t) = |νβ |ν(t)| = . j 1.2.2. We have Lij = 4π¯ hE/(|∆m2ij | c3 ). The oscillation lengths are proportional to the energy. We can use the result of question 1.3, with a conversion factor of 1000 to go from 4 MeV to 4 GeV. • For |∆m212 | c4 = 7.1 × 10−5 eV2 , we ﬁnd L12 = 140 000 km. • For |∆m223 | c4 = 2.5 × 10−3 eV2 , we ﬁnd L23 = 4 000 km. 1.2.3. We want to know the minimal distance necessary in order to observe oscillations. We assume that both mixing angles θ12 and θ23 are equal to π/4, which corresponds to maximum mixing. We saw in the ﬁrst part that if this mixing is not maximum, the visibility of the oscillations is reduced and that the distance which is necessary to observe the oscillation phenomenon is increased. By resuming the argument of the ﬁrst part, we ﬁnd that the modiﬁcation of the neutrino ﬂux of a given species is detectable beyond a distance ij such that sin2 (πij /Lij ) ≥ 0.1 i.e. ij ≥ Lij /10. This corresponds to 12 ≥ 14000 km for the oscillation resulting from the superposition 1 ↔ 2, and 23 ≥ 400 km for the oscillation resulting from the superposition 2 ↔ 3. 1.2.4. The factor of 2 between the expected muon and electron neutrino ﬂuxes comes from a simple counting. Each particle π − (resp. π + ) gives rise to a νµ , a ν¯µ and a ν¯e (resp. a νµ , a ν¯µ and a νe ). In practice, part of the muons reach the ground before decaying, which modiﬁes this ratio. Naturally, this eﬀect is taken into account in an accurate treatment of the data. 1.2.5. For an energy of 4 GeV, we have found that the minimum distance to observe the oscillation resulting from the 1 ↔ 2 superposition is 14000 km. We therefore remark that the oscillations νe νµ , corresponding to the mixing 1 ↔ 2 which we studied in the ﬁrst part cannot be observed at terrestrial distances. At such energies (4 GeV) and for evolution times corresponding at

26

1 Neutrino Oscillations

most to the diameter of the Earth (0.04 s), the energy diﬀerence E1 − E2 and the oscillations that it induces can be neglected. However, if the estimate |∆m223 | c4 > 10−3 eV2 is correct, the terrestrial distance scales allow in principle to observe oscillations resulting from 2 ↔ 3 and 1 ↔ 3 superpositions, which correspond to νµ ντ or νe ντ . 1.2.6. The angular distribution (therefore the distribution in ) observed for the νe ’s does not show any deviation from the prediction made without any oscillation. However, there is a clear indication for νµ oscillations: there is a deﬁcit of muon neutrinos coming from below, i.e. those which have had a long time to evolve. The deﬁcit in muon neutrinos is not due to the oscillation νe νµ of the ﬁrst part. Indeed, we have seen in the previous question that this oscillation is negligible at time scales of interest. The experimental data of Fig. 1.2 conﬁrm this observation. The deﬁcit in muon neutrinos coming from below is not accompanied with an increase of electron neutrinos. The eﬀect can only be due to a νµ ντ oscillation.1 No oscillation νe ντ appears in the data. In the framework of the present model, this is interpreted as the signature of a very small (if not zero) θ13 mixing angle. 1.2.7. Going back to the probability (1.11) written in question 1.4, the probability for an atmospheric muon neutrino νµ to be detected as a νµ is:

P () = 1 − sin (2θ23 ) sin 2

2

π L23

,

(1.12)

where the averaging is performed on the energy distribution of the neutrino. If we measure the neutrino ﬂux coming from the top, we have L23 , which gives Ptop = 1. If the neutrino comes from the bottom, the term sin2 (π/L23 ) averages to 1/2 and we ﬁnd: Pbottom = 1 −

1 sin2 (2θ23 ) . 2

The experimental data indicate that for −1 ≤ cos α ≤ −0.5, Pbottom = 1/2. The distribution is very ﬂat at a value of 100 events, i.e. half of the top value (200 events). We deduce that sin2 (2θ23 ) = 1, i.e. θ23 = π/4 and a maximum mixing angle for νµ ντ . The results published by Super-Kamiokande fully agree with this analysis.

1

For completeness, physicists have also examined the possibility of a “sterile” neutrino oscillation, i.e. an oscillation with a neutrino which would have no detectable interaction with matter.

1.4 Comments

27

1.4 Comments The diﬃculty of such experiments comes from the smallness of the neutrino interaction cross sections with matter. The detectors are enormous water tanks, where about ten events per day are observed (for instance ν¯e + p → e+ + n). The “accuracy” of a detector comes mainly from the statistics, i.e. the total number of events observed. In 1998, the ﬁrst undoubted observation of the oscillation ντ νµ was announced in Japan by the Super-Kamiokande experiment Fukuda Y. et al., Phys. Rev. Lett. 81, 1562 (1998)). This experiment uses a detector containing 50 000 tons of water, inside which 11 500 photomutipliers detect the Cherenkov light of the electrons or muons produced. About 60 ντ ’s were also detected, but this ﬁgure is too small to give further information. An accelerator experiment conﬁrmed the results afterwards (K2K collaboration, Phys. Rev. Lett. 90, 041801 (2003)). The KamLAND experiment is a collaboration between Japanese, American and Chinese physicists. The detector is a 1000 m3 volume ﬁlled with liquid scintillator (an organic liquid with global formula C-H). The name means KAMioka Liquid scintillator Anti-Neutrino Detector. Reference: KamLAND Collaboration, Phys. Rev. Lett. 90, 021802 (2003); see also http:/kamland.lbl.gov/. Very many experimental results come from solar neutrinos, which we have not dealt with here. This problem is extremely important, but somewhat too complex for our purpose. The pioneering work is due to Davis in his celebrated paper of 1964 (R. Davis Jr., Phys. Rev Lett. 13, 303 (1964)). Davis operated on a 37 Cl perchlorethylene detector and counted the number of 37 Ar atoms produced. In 25 years, his overall statistics has been 2200 events, i.e. one atom every 3 days! In 1991, the SAGE experiment done with Gallium conﬁrmed the deﬁcit (A. I. Abasov et al., Phys. Rev Lett. 67, 3332 (1991) and J. N. Abdurashitov et al., Phys. Rev Lett. 83, 4686 (1999)). In 1992, the GALLEX experiment, using a Gallium target in the Gran Sasso, also conﬁrmed the solar neutrino deﬁcit (P. Anselmann et al., Phys. Lett. B285, 376 (1992)). In 2001 the Sudbury Neutrino Observatory (SNO) gave decisive experimental results on solar neutrinos (Q.R. Ahmad et al., Phys. Rev. Lett. 87, 071307 (2001) and 89, 011301 (2002); see also M.B. Smy, Mod. Phys. Lett. A 17, 2163 (2002)). The 2002 Nobel prize for physics was awarded to Raymond Davis Jr. and Masatoshi Koshiba, who are the pioneers of this chapter of neutrino physics.

11 The EPR Problem and Bell’s Inequality

When a quantum system possesses more than one degree of freedom, the associated Hilbert space is a tensor product of the spaces associated to each degree of freedom. This structure leads to speciﬁc properties of quantum mechanics, whose paradoxical character has been pointed out by Einstein, Podolsky and Rosen. Here we study an example of such a situation, by considering entangled states for the spins of two particles. The system under consideration is a hydrogen atom which is dissociated into an electron and a proton. We consider the spin states of these two particles when they have left the dissociation region and are located in geometrically distinct regions, e.g. a few meters from one another. They are then free particles whose spin states do not evolve.

11.1 The Electron Spin Consider a unit vector uϕ in the (z, x) plane: uϕ = cos ϕ uz + sin ϕ ux , where ˆ e .uϕ the uz and ux are unit vectors along the z and x axes. We note Sˆeϕ = S component of the electron spin along the uϕ axis. 11.1.1. What are the eigenvalues of Sˆeϕ ? 11.1.2. We denote the eigenvectors of Sˆeϕ by |e : + ϕ and |e : − ϕ which, in the limit ϕ = 0, reduce respectively to the eigenvectors |e : + and |e : − of Sˆez . Express |e : + ϕ and |e : − ϕ in terms of |e : + and |e : −. 11.1.3. Assume the electron is emitted in the state |e : + ϕ. One measures the component Sˆeα of the spin along the direction uα = cos α uz + sin α ux . What is the probability P+ (α) of ﬁnding the electron in the state |e : + α? What is the expectation value Sˆeα in the spin state |e : + ϕ?

100

11 The EPR Problem and Bell’s Inequality

11.2 Correlations Between the Two Spins We ﬁrst assume that, after the dissociation, the electron–proton system is in the factorized spin state |e : + ϕ ⊗ |p : − ϕ. We recall that if |u1 and |u2 are vectors of E, and |v1 and |v2 of F , if ˆ act |u ⊗ |v belongs to the tensor product G = E ⊗ F , and if Aˆ and B ˆ acting in G, one has: respectively in E and F , Cˆ = Aˆ ⊗ B ˆ 1 v2 |B|v ˆ 1 . u2 | ⊗ v2 | Cˆ |u1 ⊗ |v1 = u2 |A|u

11.2.1. What is the probability P+ (α) of ﬁnding +¯ h/2 when measuring the component Sˆeα of the electron spin in this state? Having found this value, what is the state of the system after the measurement? Is the proton spin state aﬀected by the measurement of the electron spin? 11.2.2. Calculate the expectation values Sˆeα and Sˆpβ of the components of the electron and the proton spins along axes deﬁned respectively by uα and uβ = cos β uz + sin β ux . 11.2.3. The correlation coeﬃcient between the two spins E(α, β) is deﬁned as Sˆeα ⊗ Sˆpβ − Sˆeα Sˆpβ E(α, β) = . (11.1) 1/2 2 2 ˆ ˆ Seα Spβ Calculate E(α, β) in the state under consideration.

11.3 Correlations in the Singlet State We now assume that, after the dissociation, the two particles are in the singlet spin state: 1 |Ψs = √ |e : + ⊗ |p : − − |e : − ⊗ |p : + . (11.2) 2 11.3.1. One measures the component Sˆeα of the electron spin along the direction uα . Give the possible results and their probabilities. 11.3.2. Suppose the result of this measurement is +¯h/2. Later on, one measures the component Sˆpβ of the proton spin along the direction uβ . Here again give the possible results and their probabilities. 11.3.3. Would one have the same probabilities if the proton spin had been measured before the electron spin? Why was this result shocking for Einstein who claimed that “the real states of two spatially separated objects must be independent of one another”?

11.4 A Simple Hidden Variable Model

101

11.3.4. Calculate the expectation values Sˆeα and Sˆpβ of the electron and the proton spin components if the system is in the singlet state (11.2). 11.3.5. Calculate E(α, β) in the singlet state.

11.4 A Simple Hidden Variable Model For Einstein and several other physicists, the solution to the “paradox” uncovered in the previous section could come from the fact that the states of quantum mechanics, in particular the singlet state (11.2), provide an incomplete description of reality. A “complete” theory (for predicting spin measurements, in the present case) should incorporate additional variables or parameters, whose knowledge would render measurements independent for two spatially separated objects. However, present experiments cannot determine the values of these parameters, which are therefore called “hidden variables”. The experimental result should then consist in some averaging over these unknown parameters. In the case of interest, a very simpliﬁed example of such a theory is the following. We assume that, after each dissociation, the system is in a factorized state |e : +ϕ ⊗ |p : −ϕ, but that the direction ϕ varies from one event to the other. In this case, ϕ is the hidden variable. We assume that all directions ϕ are equally probable, i.e. the probability density that the decay occurs with direction ϕ is uniform and equal to 1/2π. Owing to this ignorance of the value of ϕ, the expectation value of an observable Aˆ is now deﬁned to be:

ˆ = A

1 2π

2π

e : + ϕ| ⊗ p : − ϕ| Aˆ |e : + ϕ ⊗ |p : − ϕ dϕ .

(11.3)

0

11.4.1. Using the deﬁnition (11.1) for E(α, β) and the new deﬁnition (11.3) for expectation values, calculate E(α, β) in this new theory. Compare the result with the one found using “orthodox” quantum mechanics in Sect. 3.5. 11.4.2. The ﬁrst precise experimental tests of hidden variable descriptions vs. quantum mechanics have been performed on correlated pairs of photons emitted in an atomic cascade.1 Although one is not dealing with spin-1/2 particles in this case, the physical content is basically the same as here. As an example, Fig. 11.1 presents experimental results obtained by A. Aspect and his collaborators in 1982. It gives the variation of E(α, β) as a function of the diﬀerence α − β, which is found to be the only experimentally relevant 1

The precision has now been greatly improved with the use of photon pairs produced by nonlinear splitting of ultraviolet photons (for a review, see e.g. A. Aspect, Nature, vol. 398, p. 189 (18 March 1999)).

102

11 The EPR Problem and Bell’s Inequality

quantity. Which theory, quantum mechanics or the simple hidden variable model developed above, gives a good account of the experimental data?

11.5 Bell’s Theorem and Experimental Results As proved by Bell in 1965, the disagreement between the predictions of quantum mechanics and those of hidden variable theories is actually very general when one considers correlation measurements on entangled states. We now show that the correlation results for hidden variable theories are constrained by what is known as Bell’s inequality, which, however, can be violated by quantum mechanics in speciﬁc experimental conﬁgurations. Consider a hidden variable theory, whose result consists in two functions A(λ, uα ) and B(λ, uβ ) giving respectively the results of the electron and proton spin measurements. Each of these two functions takes only the two values h/2 and −¯h/2. It depends on the value of the hidden variable λ for the con¯ sidered electron–proton pair. The nature of this hidden variable need not be further speciﬁed for the proof of Bell’s theorem. The result A of course depends on the axis uα chosen for the measurement of the electron spin, but it does not depend on the axis uβ . Similarly B does not depend on uα . This locality hypothesis is essential for the following discussion.

Fig. 11.1. Measured variation of E(α, β) as a function of α − β. The vertical bars represent the experimental error bars

11.6 Solutions

103

11.5.1. Give the correlation coeﬃcient E(α, β) for a hidden variable theory in terms of the functions A and B and the (unknown) distribution law P (λ) for the hidden variable λ. 11.5.2. Show that for any set uα , uα , uβ , uβ , one has A(λ, uα ) B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) − A(λ, uα ) B(λ, uβ ) = ±

¯2 h . 2

(11.4)

11.5.3. We deﬁne the quantity S as S = E(α, β) + E(α, β ) + E(α , β ) − E(α , β) . Derive Bell’s inequality |S| ≤ 2 . 11.5.4. Consider the particular case α − β = β − α = α − β = π/4, and compare the predictions of quantum mechanics with the constraint imposed by Bell’s inequality. 11.5.5. The experimental results obtained by A. Aspect et al. are E(α, β) = −0.66 (±0.04) for α−β = π/4 and E(α, β) = +0.68 (±0.03) for α−β = 3π/4. Is a description of these experimental results by a local hidden variable theory possible? Are these results compatible with quantum mechanics?

11.6 Solutions Section 11.1: The Electron Spin 11.1.1. In the eigenbasis |e : ± of Sˆez , the matrix of Sˆeϕ is

h cos ϕ ¯ sin ϕ . 2 sin ϕ − cos ϕ The eigenvalues of this operator are +¯h/2 and −¯ h/2. 11.1.2. The corresponding eigenvectors are ϕ ϕ |e : + ϕ = cos |e : + + sin |e : − 2 2 ϕ ϕ |e : − ϕ = − sin |e : + + cos |e : −. 2 2 11.1.3. The probability amplitude is e : + α|e : + ϕ = cos((ϕ − α)/2) and the probability P+ (α) = cos2 ((ϕ − α)/2). Similarly P− (α) = sin2 ((ϕ − α)/2), and the expectation value is, ﬁnally, Sˆeα =

¯ h cos (ϕ − α) . 2

104

11 The EPR Problem and Bell’s Inequality

Section 11.2: Correlations Between the Two Spins 11.2.1. The projector on the eigenstate |e : + α, corresponding to the measured value, is |e : + α e : + α| ⊗ Iˆp , where Iˆp is the identity operator on the proton states. Therefore P+ (α) = |e : + α|e : + ϕ|2 = cos2

ϕ−α , 2

and the state after measurement is |e : + α ⊗ |p : − ϕ. The proton spin is not aﬀected, because the initial state is factorized (all probability laws are factorized). 11.2.2. One has Sˆeα =

h ¯ 2

cos (ϕ − α) and Sˆpβ = − h¯2 cos (ϕ − β).

11.2.3. By deﬁnition, one has: ¯2 ˆ h 2 Sˆeα Ie = 4

and

¯2 ˆ h 2 Ip Sˆpβ = 4

and Sˆeα ⊗ Sˆpβ = e : + ϕ|Sˆeα |e : + ϕ p : − ϕ|Sˆpβ |p : − ϕ =−

¯2 h cos(ϕ − α) cos(ϕ − β) . 4

Therefore E(α, β) = 0. This just reﬂects the fact that in a factorized state, the two spin variables are independent. Section 11.3: Correlations in the Singlet State 11.3.1. There are two possible values: h/2, corresponding to the projector |e : + α e : + α| ⊗ Iˆp , and ¯ −¯ h/2, corresponding to the projector |e : − α e : − α| ⊗ Iˆp . Therefore, the probabilities are 1 2 2 P+ (α) = |e : + α|e : +| + |e : + α|e : −| = 1/2 2 and similarly P− (α) = 1/2. This result is a consequence of the rotational invariance of the singlet state. 11.3.2. The state after the measurement of the electron spin, yielding the result +¯ h/2, is cos

α α |e : + α ⊗ |p : − − sin |e : + α ⊗ |p : + = |e : + α ⊗ |p : − α . 2 2

This simple result is also a consequence of the rotational invariance of the singlet state, which can be written as

11.6 Solutions

105

1 |Ψs = √ (|e : + α ⊗ |p : − α − |e : − α ⊗ |p : + α) . 2 Now the two possible results for the measurement of the proton spin ±¯ h/2 have probabilities P+ (β) = sin2

α−β 2

P− (β) = cos2

α−β . 2

11.3.3. If one had measured Sˆpβ ﬁrst, one would have found P+ (β) = P− (β) = 1/2. The fact that a measurement on the electron aﬀects the probabilities for the results of a measurement on the proton, although the two particles are spatially separated, is in contradiction with Einstein’s assertion, or belief. This is the starting point of the Einstein–Podolsky–Rosen paradox. Quantum mechanics is not a local theory as far as measurement is concerned. Note, however, that this non-locality does not allow the instantaneous transmission of information. From a single measurement of the proton spin, one cannot determine whether the electron spin has been previously measured. It is only when, for a series of experiments, the results of the measurements on the electron and the proton are later compared, that one can ﬁnd this non-local character of quantum mechanics. 11.3.4. Individually, the expectation values vanish, since one does not worry about the other variable: Sˆeα = Sˆpβ = 0 . 11.3.5. However, the spins are correlated and we have

α−β α−β h2 ¯ − cos2 Sˆeα ⊗ Sˆpβ = sin2 4 2 2 and therefore

E(α, β) = − cos(α − β).

Section 11.4: A Simple Hidden Variable Model 11.4.1. Using the results of Sect. 2, we have: dϕ h ¯ cos(ϕ − α) =0 Sˆeα = 2 2π and similarly Sˆpβ = 0. We also obtain dϕ ¯2 h cos(ϕ − α) cos(ϕ − β) 4 2π 2 h ¯ = − cos(α − β) . 8

Sˆeα ⊗ Sˆpβ = −

106

11 The EPR Problem and Bell’s Inequality

Therefore, in this simple hidden variable model, 1 E(α, β) = − cos(α − β). 2 In such a model, one ﬁnds a non-vanishing correlation coeﬃcient, which is an interesting observation. Even more interesting is that this correlation is smaller than the prediction of quantum mechanics by a factor 2. 11.4.2. The experimental points agree with the predictions of quantum mechanics, and undoubtedly disagree with the results of the particular hidden variable model we have considered. We must however point out that the data given in the text is not the actual measured data. The “true” results are shown in Fig. 11.2, where the error bars correspond only to statistical errors. The diﬀerence from theory (i.e. quantum mechanics) is due to systematic errors, mainly the acceptance of the detectors.

Fig. 11.2. Actual experimental variation of E(α, β) as a function of α − β

Section 11.5: Bell’s Theorem and Experimental Results 11.5.1. In the framework of a hidden variable theory, the correlation coeﬃcient is 4 P (λ) A(λ, uα ) B(λ, uβ ) dλ , E(α, β) = 2 h ¯ where P (λ) is the (unknown) distribution law for the variable λ, with P (λ) > 0 ∀λ and P (λ) dλ = 1 .

11.6 Solutions

107

Note that we assume here that the hidden variable theory reproduces the one-operator averages found for the singlet state: Spβ = P (λ) B(λ, uβ ) dλ = 0. Seα = P (λ) A(λ, uα ) dλ = 0 If this was not the case, such a hidden variable theory should clearly be rejected since it would not reproduce a well established experimental result. 11.5.2. The quantity of interest can be written:

A(λ, uα ) B(λ, uβ ) + B(λ, uβ ) + A(λ, uα ) B(λ, uβ ) − B(λ, uβ ) . h/2. The two quantities B(λ, uβ ) and B(λ, uβ ) can take only the two values ±¯ Therefore one has either h B(λ, uβ ) + B(λ, uβ ) = ±¯ or

B(λ, uβ ) + B(λ, uβ ) = 0

B(λ, uβ ) − B(λ, uβ ) = 0 B(λ, uβ ) − B(λ, uβ ) = ±¯ h,

¯ /2. hence the result, since |A(λ, uα )| = |A(λ, uβ )| = h 11.5.3. We multiply the result (11.4) by P (λ) and integrate over λ. Bell’s inequality follows immediately. 11.5.4. The quantum mechanical result for S is SQ = − cos(α − β) − cos(α − β ) − cos(α − β ) + cos(α − β) . In general, if we set θ1 = α − β, θ2 = β − α, θ3 = α − β , we can look for the extrema of f (θ1 , θ2 , θ3 ) = cos(θ1 + θ2 + θ3 ) − (cos θ1 + cos θ2 + cos θ3 ) . The extrema correspond to θ1 = θ2 = θ3 and sin θ1 = sin 3θ1 , whose solutions between 0 and π are θ1 = 0, π/4, 3π/4, π . Deﬁning the √ function g(θ1 )√= −3 cos θ1 + cos 3θ1 we have: g(0) = −2, g(π/4) = −2 2, g(3π/4) = 2 2, g(π) = 2. We have represented the variation of g(θ) in Fig. 11.3. The shaded areas correspond to results which cannot be explained by hidden variable√theories. In particular, for α − β = β − α = α − β = π/4, we get SQ = −2 2, which clearly violates Bell’s inequality. This system constitutes therefore a test of the predictions of quantum mechanics vs. any local hidden variable theory. 11.5.5. The numbers given in the text lead to |3E(π/4) − E(3π/4)| = √ 2.66 (±0.15) in excellent agreement with quantum mechanics (2 2) but incompatible with hidden variable theories. As in the previous question, the actual measurements were in fact E(π/4) = −0.62 (±0.04), E(3π/4) = 0.60 (±0.03), therefore |3E(π/4) − E(3π/4)| =

108

11 The EPR Problem and Bell’s Inequality

2.46 (±0.15) which violates unquestionably Bell’s inequality, and is consistent with the quantum mechanical prediction. It is therefore not possible to ﬁnd a local hidden variable theory which gives a good account of experiment.

Fig. 11.3. Variation of g(θ), as deﬁned in the text

Section 11.6: References A. Einstein, B. Podolsky, and N. Rosen, Phys. Rev. 47, 777 (1935). J.S. Bell, Physics 1, 195 (1964); see also J. Bell, Speakable and unspeakable in quantum mechanics, Cambridge University Press, Cambridge (1993). The experimental data shown here are taken from: A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1982); A. Aspect, J. Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804 (1982).

18 Exact Results for the Three-Body Problem

The three-body problem is a famous question of mechanics. Henri Poincar´e was the ﬁrst to prove exact properties, and this contributed to his celebrity. The purpose of this chapter is to derive some rigorous results for the threebody problem in quantum mechanics. Here we are interested in obtaining rigorous lower bounds on three-body ground state energies. Upper bounds are easier to obtain by variational calculations. We will see that our lower bounds are actually quite close to the exact answers, to which they provide useful approximations.

18.1 The Two-Body Problem Consider a system of two particles with equal masses m and momenta p1 and p2 , interacting via a potential V (r12 ) where r12 = |r 1 − r 2 |. ˆ of the system. Let P = p1 + p2 and p = 18.1.1. Write the Hamiltonian H (p1 − p2 )/2 be the total and relative momentum. ˆ 12 Hamiltonians by writing ˆ cm and the relative H Separate the center of mass H ˆ as: H ˆ =H ˆ cm + H ˆ 12 , H

Hcm

ˆ2 P , = 2M

ˆ2 ˆ 12 = p H + V (ˆ r12 ) , 2µ

(18.1)

where M = 2m is the total mass of the system. Give the value of the reduced mass µ in terms of m. ˆ 12 . Give the ex18.1.2. We denote by E (2) (µ) the ground state energy of H (2) 2 pression for E (µ) in the two cases V (r) = −b /r and V (r) = κr2 /2.

186

18 Exact Results for the Three-Body Problem

18.2 The Variational Method ˆ and {En } the Let {|n} be the orthonormal eigenstates of a Hamiltonian H ordered sequence of its corresponding eigenvalues: E0 < E1 < E2 < · · · . ˆ 18.2.1. Show that n|H|n = En . 18.2.2. Consider an arbitrary vector |ψ of the Hilbert space of the system. By expanding |ψ on the basis {|n}, prove the inequality ∀ψ ,

ˆ ψ|H|ψ ≥ E0 ψ|ψ.

(18.2)

ˆ is the Hamiltonian 18.2.3. Show that the previous result remains valid if H of a two-body subsystem and |ψ a three-body state. In order to do so, one ˆ 12 the Hamiltonian of the (1, 2) subsystem in the three-body can denote by H system of wave function ψ(r 1 , r 2 , r 3 ). One can ﬁrst consider a given value of r 3 , and then integrate the result over this variable.

18.3 Relating the Three-Body and Two-Body Sectors Consider a system of three-particles of equal masses m with pairwise interactions: V = V (r12 ) + V (r13 ) + V (r23 ) . 18.3.1. Check the identity 2

2

2

3(p21 + p22 + p23 ) = (p1 + p2 + p3 ) + (p1 − p2 ) + (p2 − p3 ) + (p3 − p1 )

2

ˆ (3) can be written as and show that the three-body Hamiltonian H ˆ (3) = H ˆ cm + H ˆ (3) , H rel

ˆ2 ˆ cm = P , H 6m

ˆ =p ˆ1 + p ˆ2 + p ˆ 3 is the total three-body momentum, and where the where P ˆ (3) is a sum of two-particle Hamiltonians of the type relative Hamiltonian H rel deﬁned in (18.1), ˆ 12 + H ˆ 23 + H ˆ 31 ˆ (3) = H H rel with a new value µ of the reduced mass. Express µ in terms of m. ˆ ij commute in general? What would 18.3.2. Do the two-body Hamiltonians H be the result if they did? ˆ (3) , and E (3) the cor18.3.3. We call |Ω the normalized ground state of H rel responding energy. Show that the three-body ground state energy is related to the ground state energy of each two-body subsystem by the inequality: E (3) ≥ 3E (2) (µ ).

(18.3)

18.5 From Mesons to Baryons in the Quark Model

187

18.3.4. Which lower bounds on the three-body ground-state energy E (3) does one obtain in the two cases V (r) = −b2 /r and V (r) = κr2 /2? In the ﬁrst case, the exact result, which can be obtained numerically, is E (3) h2 . How does this compare with the bound (18.3)? −1.067 mb4 /¯

18.4 The Three-Body Harmonic Oscillator The three-body problem can be solved exactly in the case of harmonic interactions V (r) = κr2 /2. In order to do this, we introduce the Jacobi variables: √ √ √ ˆ 2 = (2ˆ ˆ 3 = (ˆ ˆ 1 = (ˆ r 1 − rˆ 2 )/ 2, R r 3 − rˆ 1 − rˆ 2 )/ 6, R r 1 + rˆ 2 + rˆ 3 )/ 3 R √ √ √ ˆ = (2ˆ ˆ = (ˆ ˆ = (ˆ ˆ 2 )/ 2, Q ˆ1 − p ˆ 2 )/ 6, Q ˆ2 + p ˆ 3 )/ 3 . p1 − p p3 − p p1 + p Q 1 2 3 ˆα 18.4.1. What are the commutation relations between the components R j ˆ j and Q ˆ k , (α = 1, 2, 3, and β = 1, 2, 3)? ˆ β of R and Q k 18.4.2. Check that one has Q21 + Q22 + Q23 = p21 + p22 + p23 , and: 2

2

2

3(R12 + R22 ) = (r 1 − r 2 ) + (r 2 − r 3 ) + (r 3 − r 1 ) . 18.4.3. Rewrite the three-body Hamiltonian in terms of these variables for a harmonic two-body interaction V (r) = κr2 /2. Derive the three-body ground state energy from the result. Show that the inequality (18.3) is saturated, i.e. the bound (18.3) coincides with the exact result in that case. Do you think that the bound (18.3), which is valid for any potential, can be improved without further specifying the potential?

18.5 From Mesons to Baryons in the Quark Model In elementary particle physics, the previous results are of particular interest since mesons are bound states of two quarks, whereas baryons, such as the proton, are bound states of three quarks. Furthermore, it is an empirical observation that the spectroscopy of mesons and baryons is very well accounted for by non-relativistic potential models for systems of quarks. The φ meson, for instance, is a bound state of a strange quark s and its antiquark s¯, both of same mass ms . The mass mφ is given by mφ = 2ms + E (2) (µ)/c2 where µ = ms /2, c is the velocity of light, and E (2) is the ground state energy of the s¯ s system which is bound by a potential Vqq¯(r). The Ω − baryon is made of three strange quarks. Its mass is given by MΩ = 3ms + E (3) /c2 , where E (3) is the ground state energy of the three s quarks, which interact pairwise through a two-body potential Vqq (r). These potentials are related very simply to each other by

188

18 Exact Results for the Three-Body Problem

Vqq (r) =

1 Vqq¯(r) . 2

It is a remarkable property, called ﬂavor independence, that these potentials are the same for all types of quarks. 18.5.1. Following a procedure similar to that of Sect. 3, show that E (3) ≥ (3/2)E (2) (µ ); express µ in terms of µ = ms /2. 18.5.2. Consider the potential Vqq¯(r) = g ln(r/r0 ), and the two-body Hamilˆ (2) (˜ ˆ (2) (µ) and H µ) corresponding to the same potential but diﬀerent tonians H ˆ (2) (µ) + C, ˆ (2) (˜ µ) into H reduced masses µ and µ ˜. By rescaling r, transform H where C is a constant. (2) ˆ (2) (µ) Calculate the value of C and show that the eigenvalues En (µ) of H (2) ˆ (2) (˜ and En (˜ µ) of H µ) are related by the simple formula µ) = En(2) (µ) + En(2) (˜

g µ ln . 2 µ ˜

18.5.3. A striking characteristic of the level spacings in quark–antiquark systems is that these spacings are approximately independent of the nature of the quarks under consideration, therefore independent of the quark masses. Why does this justify the form of the above potential Vqq¯(r) = g ln(r/r0 )? 18.5.4. Show that the following relation holds between the Ω − and φ masses MΩ and mφ : 3 M Ω ≥ mφ + a 2 and express the constant a in terms of the coupling constant g. 18.5.5. The observed masses are mφ = 1019 MeV/c2 and MΩ = 1672 MeV/c2 . The coupling constant is g = 650 MeV. Test the inequality with these data.

18.6 Solutions Section 18.1: The Two-Body Problem 18.1.1. The two-body Hamiltonian is ˆ 21 ˆ2 p ˆ = p H + 2 + Vˆ (r12 ) . 2m 2m The center of mass motion can be separated as usual: ˆ2 ˆ2 ˆ = P +p H + Vˆ (r12 ) , 2M 2µ where M = 2m and µ = m/2 are respectively the total mass and the reduced mass of the system.

18.6 Solutions

189

18.1.2. For a Coulomb-type interaction V (r) = −b2 /r, we get E (2) (µ) = −

µb4 . 2¯ h2

For a harmonic interaction V (r) = κr2 /2, we get κ 3 (2) E (µ) = ¯ h . 2 µ Section 18.2: The Variational Method ˆ 18.2.1. By deﬁnition, n|H|n = En n|n = En . 18.2.2.Since {|n} is a basis of the Hilbert space, |ψcan be expanded as |ψ = cn |n, and of its norm is ψ|ψ = |cn |2 . We therefore the square 2 ˆ have ψ|H|ψ = En |cn | . If we simply write ˆ (En − E0 )|cn |2 , ψ|H|ψ − E0 ψ|ψ = we obtain, since En ≥ E0 and |cn |2 ≥ 0: ˆ ψ|H|ψ ≥ E0 ψ|ψ . ˆ =H ˆ 12 , for ﬁxed r 3 , ψ(r 1 , r 2 , r 3 ) can be considered as a non18.2.3. If H normalized two-body wave function. Therefore ∗ ˆ 12 ψ(r 1 , r 2 , r 3 ) d3 r1 d3 r2 ψ (r 1 , r 2 , r 3 ) H ≥ E0 |ψ(r 1 , r 2 , r 3 )|2 d3 r1 d3 r2 . By integrating this inequality over r 3 , one obtains the desired result. Section 18.3: Relating the Three-Body and Two-Body Sectors 18.3.1. The identity is obvious, since the crossed terms vanish on the rightˆ 2 /(6m) + H ˆ =P ˆ 12 + H ˆ 23 + H ˆ 31 , with hand side. Therefore H

2

#2 " ˆj ˆ j /2 ˆi − p ˆi − p p p ˆ ˆ Hij = + V (rij ) = + Vˆ (rij ) (18.4) 6m 2µ with a reduced mass µ = 3m/4. ˆ 12 and H ˆ 23 do not commute; for instance p ˆ1 − p ˆ 2 does 18.3.2. Obviously, H ˆ not commute with V (r23 ). If they did, the three-body energies would just be the sum of two-body energies as calculated with a reduced mass µ = 3m/4, and the solution of the three-body problem would be simple.

190

18 Exact Results for the Three-Body Problem

ˆ (3) |Ω = Ω|H ˆ ij |Ω. However, owing 18.3.3. By deﬁnition, E (3) = Ω|H rel ˆ ij |Ω ≥ E (2) (µ ), so that to the results of questions 2.2 and 2.3, we have Ω|H E (3) ≥ 3E (2) (µ )

with

µ = 3m/4 .

18.3.4. For a Coulomb-type potential, we obtain E (3) ≥ −

3 µ b4 9 mb4 =− , 2 2 ¯ 8 ¯h2 h

which deviates by only 6% from the exact answer −1.067 mb4 /¯ h2 . In the harmonic case, we obtain: √ κ κ 3 (3) h . E ≥3 ¯ = 3 3 ¯h 2 µ m Section 18.4: The Three-Body Harmonic Oscillator 18.4.1. One easily veriﬁes that Jacobi variables satisfy canonical commutation relations: ˆα, Q ˆ β ] = i¯ [R h δjk δαβ . j k 18.4.2. These relations are a simple algebraic exercise. 18.4.3. We ﬁnd ˆ2 ˆ2 ˆ2 ˆ = Q1 + 3 κR ˆ 2 + Q2 + 3 κR ˆ 2 + Q3 = H ˆ1 + H ˆ2 + H ˆ cm , H 1 2 2m 2 2m 2 2m ˆ cm = Q ˆ 2 /(2m) = Pˆ 2 /(6m) is the center of mass Hamiltonian. The where H 3 ˆ 2 , and H ˆ cm commute. The ground state energy (with ˆ 1, H three Hamiltonians H the center of mass at rest) is therefore √ 3κ κ 3 (3) h = 3 3 ¯h , E =2 ¯ 2 m m which coincides with the lower bound obtained in question 3.4. The bound is therefore saturated if the interaction is harmonic. In order to improve the bound, one must further specify the interaction. Actually, the bound is saturated if and only if the interaction potential is harmonic. Indeed the variational inequality we use becomes an equality if and only if the wave function coincides with the exact ground state wave function. Owing to the particular symmetry of quadratic forms, the Jacobi variables guarantee that this happens in the harmonic case. The property ceases to be true for any other potential.

18.6 Solutions

191

Section 18.5: From Mesons to Baryons in the Quark Model 18.5.1. The s¯ s relative Hamiltonian is 2 ˆ (2) = pˆ + Vqq¯(ˆ H r) . ms

The sss relative Hamiltonian is (cf. Sect. 3): (ˆ pi − pˆj )2 1 (3) ˆ + Vqq¯(ˆ rij ) H = 6ms 2 i<j

pi − pˆj )2 1 (ˆ = + Vqq¯(ˆ rij ) . 2 i<j 3ms

(18.5) (18.6)

Therefore, ˆ (3) = 2H

2

ˆ ij H

pi − pˆj ) /2) ˆ ij = ((ˆ H + Vqq¯(ˆ rij ) 2µ

with

i<j

with µ = 3ms /8 = 3µ/4. From this relation we deduce the inequality: 2 E (3) ≥ 3 E (2) (µ )

µ = 3µ/4 .

with

18.5.2. With the rescaling r → αr, one obtains: ˆ (2) (˜ H µ) = The choice α =

r pˆ2 + g ln + g ln α . 2 2α µ ˜ r0

ˆ (2) (˜ ˆ (2) (µ) + g ln α so that µ/˜ µ leads to H µ) = H En(2) (˜ µ) = En(2) (µ) +

g µ ln . 2 µ ˜

18.5.3. In a logarithmic potential, the level spacing is independent of the mass. This is a remarkable feature of the observed spectra, at least for heavy quarks, and justiﬁes the investigation of the logarithmic potential. Amazingly enough, this empirical prescription works quite well for light quarks, although one might expect that a relativistic treatment is necessary. 18.5.4. The binding energies satisfy

3 g 4 E (3) ≥ E (2) + ln 2 2 3 with MΩ = 3ms +

E (3) c2

We therefore obtain MΩ ≥

mφ = 2ms +

3 3g 4 mφ + 2 ln . 2 4c 3

E (2) . c2

192

18 Exact Results for the Three-Body Problem

18.5.5. For g = 650 MeV and a = 140 MeV/c2 , we obtain MΩ c2 = 1672 MeV ≥ 1669 MeV , which is remarkably accurate. Actually, the quark–quark potential is only logarithmic at distances smaller than 1 fm, which corresponds to the φ mean square radius. At larger distances, it grows more rapidly (linearly). Such inequalities are quite useful in practice for deciding what choice to make for the potential and for its domain of validity. The generalization of such inequalities can be found in the literature quoted below. They are useful in a variety of physical problems. References J.-L. Basdevant, J.-M. Richard, and A. Martin, Nuclear Physics B343, 60, 69 (1990). J-L. Basdevant, J.-M. Richard, A. Martin, and Tai Tsun Wu, Nuclear Physics B393, 111 (1993).

Author Index

Abasov, A.I. et al., 27 Abdurashitov, J.N. et al., 27 Ahmad, Q. R. et al., 27 Anderson, M.H., 202 Andrews, M.R., 202 Anselmann, P. et al., 27 Ashkin, A., 276 Aspect, A., 101, 108 Barnett, S.M., 167 Basdevant, J.-L., 192 Bastard, G., VII Bell, J.S., 99, 108 Ben Dahan, M., 285 Berkhout, J., 265 Berko, S., 80 Berlinsky, A.J., 265 Biraben, F., 82 Bjorkholm, J.E., 276 Bouchoule, I., 285 Bradley, C., 202 Brune, M., 119, 145 Cable, A., 276 Castin, Y., VII, 285 Chazalviel, J.-N., VII Chazalviel, J.N., 239 Chu, S., 80, 276 Clairon, A., 36 Cohen-Tannoudji, C., 276 Collela, A.R., 45 Corben, H.C., 80 Cornell, E.A., 202 Courty, J.-M., VII

Crane, H.R., 66 Dalibard, J., 108 Davis R., 27 DeBenedetti, S., 80 Dehmelt, H., 276 Delande, D., VII, 82 Dorner, B., 213 Doyle, J.M., 265 Dress, W.D., 53 Dreyer, J., 119, 145 Durfee, D.S., 202 Einstein, A., 99, 100, 108 Ensher, J.R., 202 Equer, B., VII Fischer, M., 36 Fukuda Y., 27 Gabrielse, G., 182 GALLEX Collaboration, 27 Gay, J.-C., 82 Gillet, V., VII Grangier, P., VII, 108, 153, 167 Greenberger, D., 45 Greene, G.L., 53 Grynberg, G., VII H¨ ansch, T.W., 276 Hagley, E., 119, 145 Hameau, S., 229 Haroche, S., 119, 145 Hollberg, L., 276

288

Author Index

Hulet, R.G., 202

Prodan, J.V., 276

Jacquet, F., VII Jolicoeur, T., VII K2K collaboration, 27 KamLAND collaboration, 27 Ketterle, W., 202 Kinoshita, T., 80 Koshiba M., 27 Kurn, D.M., 202

Raimond, J.-M., 119, 145 Raizen, M., 285 Ramsey, N.F., 47, 53, 156 Reichel, J., 285 Rich, J., VII Richard, J.-M., 192 Roger, G., 108 Rosen, N., 99, 108 Roug´e, A., VII

Maali, A., 119, 145 Maitre, X., 119 Maleki, L., 36 Mampe, W., 53 Marion, H., 36 Martin, A., 192 Matthews, M.R., 202 Metcalf, H., 276 Mewes, M.-O., 202 Miller, P.D., 53 Mills, A.P., 80

Sackett, C.A., 202 SAGE Collaboration, 27 Salomon, C., 36, 285 Santarelli, G., 36 Schawlow. A., 276 Schmidt-Kaler, F., 145 Schr¨ odinger, 109 Smith, J.H., 166 Smith, K.F., 53, 167 Smy, M.B., 27 Stoler, D., 119

Neuman, J. von, 147 Niu, Q., 285

Tjoelker, R.L., 36

Overhauser, A.W., 37, 45

van Druten, N.J., 202 von Neuman, J., 147

Peik, E., 285 Peil, S., 182 Pendlebury, J.M., 53 Pendleton, H.N., 80 Penent, F., 82 Perrin, P., 53 Phillips, W.D., 276 Platt, J.R., 235 Podolsky, B., 99, 108 Poizat, J.P., 153, 167 Prestage, J.D., 36

Weinheimer et al., 72 Werner, S.A., 45 Wieman, C.E., 202 Wilkinson, D.T., 66 Wineland, D., 276 Wu, Tai Tsun, 192 Wunderlich, C., 119 Yurke, B., 119 Zimmerman, D.S., 265

Subject Index

absorption, 239, 272 addition of angular momenta, 7, 29 adiabatic theorem, 278 alakali atom, 29 angular momenta addition, 7, 29 angular momentum, 6 annihilation, 75, 132 annihilation and creation operators, 109 antineutrino, 69 antinode, 234 antiparticle, 17 antiquark, 187 atmospheric neutrinos, 20 atomic clocks, 29 atomic fountain, 31 β decay, 17, 69 Band structure, 277 baryon, 187 Bell’s inequality, 99, 102, 107 Bell’s theorem, 102 Bethe-Bloch formula, 94 Bloch index, 278 Bloch oscillations, 277, 278, 284 Bloch theorem, 278 blue shift, 233 Bohr radius, 69, 82, 240 Born approximation, 12 Bose–Einstein condensate, 193, 195 boson, 10 Bragg angle, 37 Bragg reﬂection, 285

Brillouin zone, 281, 284 broad line condition, 273 cat paradox, 109 cesium, 29, 32 cesium atom, 283 cesium atoms, 279 • CH3 (methyl), 239 CH3 −• COH−COO− , 240 chain of coupled spins, 203 chain of molecules, 204 coherence properties, 193 collision, 12 condensate, 193 conductor, 285 cooling time, 273, 274 correlated pairs of spins, 122 correlation coeﬃcient, 100 Coulomb correlations, 229 Coulomb potential, 8 creation, 132 cross section, 12 crossed Kerr eﬀect, 153 cryptography, 121 CSCO, 3 Cs Fe Br3 , 204 damped oscillation, 135 damping, 113 damping time, 137, 144 decay, 75 degeneracy, 206 density operator, 3, 247, 251, 267 diﬀraction peak, 60

290

Subject Index

Dirac distribution, 5 dispersion relation, 257 Doppler cooling, 267, 269 Doppler shift, 52 drift of fundamental constants, 32

hydrogen hyperﬁne hyperﬁne hyperﬁne hyperﬁne

Ehrenfest theorem, 4, 59, 61, 67, 139 Einstein-Podolsky-Rosen paradox, 99 electric dipole, 267 electric dipole coupling, 277 electric dipole moment, 166 electromagnetic wave, 31 electron magnetic moment anomaly, 65 electron spin, 100 electron spin resonance, 237 energy bands, 281 energy loss, 87 entangled states, 3, 99 entanglement, 3 eraser, 155 excitons, 206

identical particles, 10 interaction time, 88, 91 interference, 155, 159 ion beams, 94 ion therapy, 94 ionization energy, 69 ions in matter, 87 isotope, 69

atom, 73, 81 interaction, 29, 237, 238 splitting, 73 structure, 237 transition, 74

Jacobi variables, 187, 190 joint probability, 165 KamLAND, 19 ket, 1

G.P.S., 29, 32 gravitational ﬁeld, 37 gravitational potential, 43 gyromagnetic ratios, 237

Laboratoire Kastler Brossel, 145 Laguerre polynomial, 9 Larmor frequency, 67 Larmor precession, 47 laser cooling, 267 LEP, 18 lepton, 17 lifetime, 12 liquid helium, 255, 258 lithium atoms, 197 local hidden variable theory, 103 local probe, 245 local theory, 105 locality, 102 logarithmic potential, 191 lower bounds, 185

Hamiltonian, 2 harmonic oscillator, 7, 109, 197 harmonic potential, 193 3 He+ , 69 heavy ion therapy, 94 Heisenberg inequalities, 60 hidden variables, 99, 101 Hilbert space, 1 holes, 211 hydrocarbon ions, 231 hydrogen, 249, 255

magnetic excitations, 203 magnetic excitons, 203, 206 magnetic interaction, 29, 41 magnetic moment, 58 magnetic resonance, 155 measurement, 147, 158 measuring apparatus, 63, 159 meson, 17, 187 modes, 267 molecular ions, 231 muon, 17

factorized spin, 100 Fermi golden rule, 11 fermion, 10 ﬁeld quantization, 131 ﬁne structure constant, 32 ﬂavor, 20 form factor, 13 Fourier transform, 4 free radicals, 237

Subject Index muon spin, 245 muonium, 245 muonium in silicon, 247

proton spin, 100, 105 proton therapy, 94 pure case, 1

n photon state, 132 neutrino, 17, 253 neutrino mass, 71 neutrino masses, 18 neutrino oscillations, 17 neutron, 55, 155, 206 neutron beam, 37, 47, 55 neutron gyromagnetic ratio, 49 neutron interference, 37 neutron magnetic moment, 59 nitrogenous ions, 232 node, 234 nuclear reaction, 87 nucleus, 29, 69

QND measurement, 153, 167 quantum boxes, 215 quantum cryptography, 121, 125 quantum dots, 215 quantum eraser, 155, 159 quantum measurement, 2 quantum paths, 155, 162 quantum reﬂection, 255 quantum revival, 137 quantum superposition, 112, 113 quark masses, 188 quark model, 187 quasi-classical states, 110, 133 quasi-particles, 203

observable, 2 Operator, 2 Operator energy, 5 momentum, 5 position, 5 optical Bloch equations, 267, 271 optronics, 229 orbital angular momentum, 6 oscillation Rabi, 11 oscillation lengths, 21

Rabi oscillation, 11, 131, 143 radiation pressure, 268 radioactive, 69 Ramsey fringes, 47, 156 red shift, 233 reﬂection, 255 relaxation, 267 resonance curve, 50 revival time, 144 ripplon, 255, 257, 263, 265 rubidium, 29, 32, 135

pairs of photons, 101 Pauli hydrogen atom, 81 Pauli matrices, 40, 121, 237 Pauli principle, 231 periodic potential, 277 perturbation theory, 9 phase states, 148 phonon, 229 photon pairs, 129 pigments, 231 polyethylene, 231 positive muon, 245 positron, 73, 253 positronium, 73 potential well, 231 precession, 65

satellite, 32 scalar product, 1 scattering, 12 scattering amplitude, 13 scattering length, 14, 194 scattering state, 13 Schr¨ odinger’s cat, 109 Schr¨ odinger equation, 2 semiconductor, 229 semiconductor detectors, 89 singlet state, 76, 127 sodium atom, 268, 269 sodium condensate, 196 spatial state, 57 spherical harmonics, 6 spin, 7 spin excitation wave, 207

291

292

Subject Index

spin state, 41, 47, 57, 122, 155, 158, 159, 246 spin transitions, 238 spin–spin interaction, 204, 246 spontaneous emission, 268, 272 standing laser wave, 277 Stark eﬀect, 81 state, 1 statistical mixture, 3, 112, 113 sterile neutrino, 26 Stern–Gerlach experiment, 55, 63 sticking probability, 255, 265 stimulated emission, 272 Super-Kamiokande, 21 superconducting niobium cavity, 145 supernovae, 18 superposition principle, 1 τ lepton, 17 Thomas–Reiche–Kuhn sum rule, 89 three-body harmonic oscillator, 187 three-body problem, 185 time–energy uncertainty relation, 63 time-dependent perturbations, 11 transition amplitude, 51

triplet states, 76, 77 tritium atom, 69 tritium decay, 69 uncertainty relation, 4, 5, 50 unitary operator, 277 unpaired electron, 237 Van der Waals potential, 255 variational inequality, 190 variational method, 10, 186, 193 visible light, 233 von Neumann detector, 147 wave function, 4 wave packet projection, 2 Weak interactions, 17 width, 50, 51, 273 Wigner–Weisskopf, 271 WKB approximation, 256 Yukawa potential, 12 Zeeman eﬀect, 74, 81 Zeeman interaction, 237, 238

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